combinatorics through guided discovery · 2019-08-19 · combinatorics through guided discovery...

Kenneth P. Bogart

CombinatoricsThrough

Guided Discovery

Combinatorics Through GuidedDiscovery

Combinatorics Through GuidedDiscovery

Kenneth P. BogartDartmouth College

EditorsMitchel T. Keller

Washington & Lee University

Oscar LevinUniversity of Northern Colorado

Kent E. MorrisonAmerican Institute of Mathematics

Edition: First PreTeXt Edition

Website: http://bogart.openmathbooks.org/

© 2004 (main text); 2017 (Preface to PreTeXt Edition) Kenneth P. Bogart (maintext); Mitchel T. Keller, Oscar Levin, Kent E. Morrison (Preface to PreTeXt Edition)

Permission is granted to copy, distribute and/or modify this document under theterms of the GNU Free Documentation License, Version 1.3 or any later versionpublished by the Free Software Foundation; with no Invariant Sections, no Front-Cover Texts, and no Back-Cover Texts. A copy of the license is included in theappendix entitled “GNU Free Documentation License”.

http://bogart.openmathbooks.org/

About the Author

Kenneth Paul Bogart was born on October 6, 1943 in Cincinnati, OH. He gradu-ated from Marietta College in Ohio in 1965, and earned his Ph.D. in mathematicsfrom the California Institute of Technology in 1968. He married Ruth Tucker in1966, and they moved to Hanover in 1968 where Ken was appointed an AssistantProfessor of Mathematics. Ken remained in the job that he loved for 37 years be-ing promoted to Associate Professor in 1974, and to Full Professor in 1980. Ken’scareer was characterized by a love of mathematics and scholarship, and a passionfor teaching and mentoring at all levels within the mathematics curriculum. Hispassion for research is evidenced by over 60 journal articles and nine textbooks inhis field of combinatorics. Ken’s research covered a wide spectrum of topics withincombinatorics.

Ken’s mathematical roots were in algebra and lattice theory, and his earliestpapers developed structural results for Noether lattices. One of the main topics inhis research was partial orders, about which he wrote more than two dozen papers.This line of research started in the early 1970’s with contributions to the theory ofdimension for partial orders. A number of his papers treated applications of partialorders to the social sciences; for instance, he contributed to social choice theory byexamining the optimal way to develop a consensus based on rankings that arepartial orders. Interval orders and interval graphs played the most prominent rolein Ken’s research; his papers in this field span roughly thirty years, starting in themid 1970’s, and about half of his Ph.D. students worked in this area. Among hiscontributions in this area are the introduction and investigation of new conceptsrelated to interval orders and graphs, the development of new and simpler proofs ofkey results, and the exploration of a number of structures that are natural variationsor interesting special types of interval orders and graphs. Ken also contributed tothe theory of error-correcting codes; in particular, he constructed a class of codesfrom partial orders. He collaborated on several papers in matroid theory, to whichhe contributed valuable insights from lattice theory and geometry.

Throughout the later part of his career, Ken became increasing interested in howstudents learn mathematics. His NSF-sponsored project of "guided-discovery" incombinatorics is an element that lives on in the math department. Ken also devoted

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a great deal of time to helping revise the teaching seminar which is fundamentalpart of the mathematics graduate program at Dartmouth.

For the past nine years, Ken and Ruth spent winters in Santa Rosa, CA, wherethey loved to hike and mountain bike.

This biography is excerpted from one written by T. R. Shemanske and postedon the website of the Department of Mathematics of Dartmouth College in April2005. Shemanske wrote:

I have borrowed freely from a number of published sources (below), andam especially grateful to Professor Joe Bonin for writing about Ken’sresearch, Mary Pavone for her remarks about Ken’s involvement withWISP, and for a photo from Ruth Bogart.Joe Bonin, The George Washington University; Mary Pavone, DirectorWomen in Science Project; Press Democrat, Santa Rosa, CA (April 6,2005); The Dartmouth, (April 4, 2005)

https://math.dartmouth.edu/publicity/general/kpbogart-life.phtml

Preface

This book is an introduction to combinatorial mathematics, also known as combina-torics. The book focuses especially but not exclusively on the part of combinatoricsthat mathematicians refer to as “counting.” The book consist almost entirely ofproblems. Some of the problems are designed to lead you to think about a concept,others are designed to help you figure out a concept and state a theorem about it,while still others ask you to prove the theorem. Other problems give you a chanceto use a theorem you have proved. From time to time there is a discussion thatpulls together some of the things you have learned or introduces a new idea for youto work with. Many of the problems are designed to build up your intuition forhow combinatorial mathematics works. There are problems that some people willsolve quickly, and there are problems that will take days of thought for everyone.Probably the best way to use this book is to work on a problem until you feel you arenot making progress and then go on to the next one. Think about the problem youcouldn’t get as you do other things. The next chance you get, discuss the problemyou are stymied on with other members of the class. Often you will all feel you’vehit dead ends, but when you begin comparing notes and listening carefully to eachother, you will see more than one approach to the problem and be able to makesome progress. In fact, after comparing notes you may realize that there is morethan one way to interpret the problem. In this case your first step should be tothink together about what the problem is actually asking you to do. You may havelearned in school that for every problem you are given, there is a method that hasalready been taught to you, and you are supposed to figure out which methodapplies and apply it. That is not the case here. Based on some simplified examples,you will discover the method for yourself. Later on, you may recognize a patternthat suggests you should try to use this method again.

The point of learning from this book is that you are learning how to discoverideas and methods for yourself, not that you are learning to apply methods thatsomeone else has told you about. The problems in this book are designed to leadyou to discover for yourself and prove for yourself the main ideas of combinatorialmathematics. There is considerable evidence that this leads to deeper learning andmore understanding.

You will see that some of the problems are marked with bullets. Those are theproblems that I feel are essential to having an understanding of what comes later,whether or not it is marked by a bullet. The problems with bullets are the problemsin which the main ideas of the book are developed. Your instructor may leave outsome of these problems because he or she plans not to cover future problems thatrely on them. Many problems, in fact entire sections, are not marked in this way,

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because they use an important idea rather than developing one. Some other specialsymbols are described in what follows; a summary appears in the table below.

• essential◦ motivational material+ summary⇒ especially interesting∗ difficult· essential for this section or the next

Some problems are marked with open circles. This indicates that they aredesigned to provide motivation for, or an introduction to, the important concepts,motivation with which some students may already be familiar. You will also seethat some problems are marked with arrows. These point to problems that I thinkare particularly interesting. Some of them are also difficult, but not all are. A fewproblems that summarize ideas that have come before but aren’t really essential aremarked with a plus, and problems that are essential if you want to cover the sectionthey are in or, perhaps, the next section, are marked with a dot (a small bullet).If a problem is relevant to a much later section in an essential way, I’ve marked itwith a dot and a parenthetical note that explains where it will be essential. Finally,problems that seem unusually hard to me are marked with an asterisk. SomeI’ve marked as hard only because I think they are difficult in light of what hascome before, not because they are intrinsically difficult. In particular, some of theproblems marked as hard will not seem so hard if you come back to them after youhave finished more of the problems.

If you are taking a course, your instructor will choose problems for you to workon based on the prerequisites for and goals of the course. If you are reading thebook on your own, I recommend that you try all the problems in a section youwant to cover. Try to do the problems with bullets, but by all means don’t restrictyourself to them. Often a bulleted problem makes more sense if you have donesome of the easier motivational problems that come before it. If, after you’ve triedit, you want to skip over a problem without a bullet or circle, you should not missout on much by not doing that problem. Also, if you don’t find the problems ina section with no bullets interesting, you can skip them, understanding that youmay be skipping an entire branch of combinatorial mathematics! And no matterwhat, read the textual material that comes before, between, and immediately afterproblems you are working on!

One of the downsides of how we learn math in high school is that many of uscome to believe that if we can’t solve a problem in ten or twenty minutes, then wecan’t solve it at all. There will be problems in this book that take hours of hardthought. Many of these problems were first conceived and solved by professionalmathematicians, and they spent days or weeks on them. How can you be expectedto solve them at all then? You have a context in which to work, and even thoughsome of the problems are so open ended that you go into them without any ideaof the answer, the context and the leading examples that preceded them give youa structure to work with. That doesn’t mean you’ll get them right away, but you

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will find a real sense of satisfaction when you see what you can figure out withconcentrated thought. Besides, you can get hints!

Some of the questions will appear to be trick questions, especially when youget the answer. They are not intended as trick questions at all. Instead they aredesigned so that they don’t tell you the answer in advance. For example the answerto a question that begins “How many...” might be “none.” Or there might bejust one example (or even no examples) for a problem that asks you to find allexamples of something. So when you read a question, unless it directly tells youwhat the answer is and asks you to show it is true, don’t expect the wording ofthe problem to suggest the answer. The book isn’t designed this way to be cruel.Rather, there is evidence that the more open-ended a question is, the more deeplyyou learn from working on it. If you do go on to do mathematics later in life, theproblems that come to you from the real world or from exploring a mathematicaltopic are going to be open-ended problems because nobody will have done thembefore. Thus working on open-ended problems now should help to prepare you todo mathematics later on.

You should try to write up answers to all the problems that you work on. Ifyou claim something is true, you should explain why it is true; that is you shouldprove it. In some cases an idea is introduced before you have the tools to proveit, or the proof of something will add nothing to your understanding. In suchproblems there is a remark telling you not to bother with a proof. When you writeup a problem, remember that the instructor has to be able to “get” your ideas andunderstand exactly what you are saying. Your instructor is going to choose someof your solutions to read carefully and give you detailed feedback on. When youget this feedback, you should think it over carefully and then write the solutionagain! You may be asked not to have someone else read your solutions to someof these problems until your instructor has. This is so that the instructor can offerhelp which is aimed at your needs. On other problems it is a good idea to seekfeedback from other students. One of the best ways of learning to write clearly is tohave someone point out to you where it is hard to figure out what you mean. Thecrucial thing is to make it clear to your reader that you really want to know whereyou may have left something out, made an unclear statement, or failed to supporta statement with a proof. It is often very helpful to choose people who have not yetbecome an expert with the problems, as long as they realize it will help you mostfor them to tell you about places in your solutions they do not understand, even ifthey think it is their problem and not yours!

As you work on a problem, think about why you are doing what you are doing.Is it helping you? If your current approach doesn’t feel right, try to see why. Isthis a problem you can decompose into simpler problems? Can you see a way tomake up a simple example, even a silly one, of what the problem is asking you todo? If a problem is asking you to do something for every value of an integer n,then what happens with simple values of n like 0, 1, and 2? Don’t worry aboutmaking mistakes; it is often finding mistakes that leads mathematicians to theirbest insights. Above all, don’t worry if you can’t do a problem. Some problemsare given as soon as there is one technique you’ve learned that might help dothat problem. Later on there may be other techniques that you can bring back tothat problem to try again. The notes have been designed this way on purpose. If

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you happen to get a hard problem with the bare minimum of tools, you will haveaccomplished much. As you go along, you will see your ideas appearing againlater in other problems. On the other hand, if you don’t get the problem the firsttime through, it will be nagging at you as you work on other things, and when yousee the idea for an old problem in new work, you will know you are learning.

There are quite a few concepts that are developed in this book. Since most of theintellectual content is in the problems, it is natural that definitions of concepts willoften be within problems. When you come across an unfamiliar term in a problem,it is likely it was defined earlier. Look it up in the index, and with luck (hopefullyno luck will really be needed!) you will be able to find the definition.

Above all, this book is dedicated to the principle that doing mathematics is fun.As long as you know that some of the problems are going to require more than oneattempt before you hit on the main idea, you can relax and enjoy your successes,knowing that as you work more and more problems and share more and moreideas, problems that seemed intractable at first become a source of satisfaction lateron.

The development of this book is supported by the National Science Founda-tion. An essential part of this support is an advisory board of faculty membersfrom a wide variety of institutions who have made valuable contributions. Theyare Karen Collins, Wesleyan University, Marc Lipman, Indiana University/PurdueUniversity, Fort Wayne, Elizabeth MacMahon, Lafayette College, Fred McMorris,Illinois Institute of Technology, Mark Miller, Marietta College, Rosa Orellana, Dart-mouth College, Vic Reiner, University of Minnesota, and Lou Shapiro, HowardUniversity. The overall design and most of the problems in the appendix on ex-ponential generating functions are due to Professors Reiner and Shapiro. Anyerrors or confusing writing in that appendix are due to me! I believe the board hasmanaged both to make the book more accessible and more interesting.

Preface to PreTeXt edition

At the time of his death in 2005, Ken Bogart was working on this NSF-supportedeffort to create a combinatorics textbook that developed the key ideas of undergrad-uate combinatorics through “guided discovery”, or what many today typically callinquiry-based learning. The project was under contract with Springer-Verlag for acommercially-published print edition, but Ken’s untimely passing left the projectin an unfinished state. Bogart’s family asked the Department of Mathematics atDartmouth College, where he had spent his entire career after earning his Ph.D.from Caltech in 1968, to distribute the text freely under the GNU Free Documen-tation License. This open-source release came with some notes, however. Thosenotes, listed on the book’s Dartmouth page, were:

1. The contents of the archive are released under the terms of the gnuFree Documentation License (FDL), a copy of which is containedin the archive.

2. The contents of the archive are released in “as is” condition, whichin particular means that the state of the source files is not in agree-ment with the pdf versions of the text. A README offers someguidance.

3. Many people have already used the textbook in courses at variousuniversities throughout the country. It is the hope of the Bogartfamily that this project continues to grow to completion with theefforts of those who download this archive.

The caveat in the second note seemed to be the largest toward fulfilling the goal ofthe third, as the “official” version of the pdf had a different chapter structure thanthe LATEX source files provided. This was mostly the result of splitting a chapter intotwo and rearranging a few topics, but there were also places where problems weresplit or merged between the source version and the pdf version. The pdf versionalso came with copious hints that readers could access online, but no LATEX sourceexisted for these hints.

This PreTeXt edition of Combinatorics through Guided Discovery attempts to helpfulfill the Bogart family’s wish to see the project grow and reach a complete state.One of us (mtk) had used the official pdf to teach a combinatorics course in Winter2015 and mentioned this fact at a workshop on open source textbooks and PreTeXt(then MathBook xml) organized by the American Institute of Mathematics in thespring of 2016. This caught the attention of kem, since Combinatorics through Guided

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https://www.math.dartmouth.edu/news-resources/electronic/kpbogart/

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Discovery had been placed on the aim list of approved open source textbooks, butthere had been no success in finding someone to take on the task of updating thesource to match the pdf. The three of us came together again in May 2017 at theUniversity of Puget Sound for another workshop on open educational resourcesand agreed to cooperate to complete the conversion of this book to PreTeXt. Thefact that ol wanted to use parts of the book for his Fall 2017 class gave us themotivation required to complete the project over the summer.

For this edition, our goal has been to reproduce the text of Bogart’s final pdfas faithfullly as possible. Based on our own classroom uses, we have notes aboutproblems that could use revising, but we agreed the right first step would be tohave source files that matched what Bogart left. We have, however, correctedobvious errors along the way, which included moving the Supplementary ChapterProblems in Chapter 3 from the level of a subsection to the level of a section forconsistency with the other chapters. Footnotes may be numbered differently, as inthis edition, a footnote in the body of a problem is rendered with the problem andnumbered in a different sequence. The hints that previously were accessed by linksfrom the pdf to a Dartmouth webpage have also been included in the backmatterof the print edition as Appendix D. Links to open hints in place are available inthe HTML version, while in the print and PDF edition, the existence of a hint isindicated by “(h)” at the end of the problem (or part of a problem). David Farmerprovided invaluable assistance by automating the initial conversion of LATEX filesto PreTeXt and extracting the text of the hints from the pdf files. We then workedin parallel to compare the official pdf to what we were able to produce from thesource until they matched. Since this process could not be truly automated, wesuspect there will be some places where Bogart’s pdf and this edition do not match.We welcome reports of these through issues and pull requests on the Githubsite for the book https://github.com/OpenDiscreteMath/ibl-combinatorics/. Goingforward, we would like to see community-driven updates to further develop thetext, either by improving existing problems, adding new problems on existingtopics, or adding new topics suitable for a course based on this text. One area ofdevelopment may be to include SageMath to the text, since PreTeXt includles anumber of nice features for doing this and some of the material may benefit fromthe addition of a computer algebra system to allow more interesting calculationsthan would be feasible by hand.

An html version of this text is available at http://bogart.openmathbooks.org/. Alow-cost print edition is available for purchase online. The cost of the print editionis kept as low as possible, and any royalties received support costs associated withhosting and distributing the text. A pdf copy of print edition is also posted on thebook’s site. The pdf may provide a better experience for searching than the HTMLversion.

Mitchel T. Keller, Oscar Levin, and Kent E. MorrisonLexington, Virginia; Greeley, Colorado; and San Jose, CaliforniaDecember 2017

https://github.com/OpenDiscreteMath/ibl-combinatorics/

http://bogart.openmathbooks.org/

Contents

About the Author v

Preface vii

Preface to PreTeXt edition xi

1 What is Combinatorics? 11.1 About These Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Basic Counting Principles . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2.1 The sum and product principles . . . . . . . . . . . . . . . . . 51.2.2 Functions and directed graphs . . . . . . . . . . . . . . . . . . 91.2.3 The bĳection principle . . . . . . . . . . . . . . . . . . . . . . . 111.2.4 Counting subsets of a set . . . . . . . . . . . . . . . . . . . . . 111.2.5 Pascal’s Triangle . . . . . . . . . . . . . . . . . . . . . . . . . . 121.2.6 The quotient principle . . . . . . . . . . . . . . . . . . . . . . . 15

1.3 Some Applications of the Basic Principles . . . . . . . . . . . . . . . . 201.3.1 Lattice paths and Catalan Numbers . . . . . . . . . . . . . . . 201.3.2 The Binomial Theorem . . . . . . . . . . . . . . . . . . . . . . . 231.3.3 The pigeonhole principle . . . . . . . . . . . . . . . . . . . . . 251.3.4 Ramsey Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . 26

1.4 Supplementary Chapter Problems . . . . . . . . . . . . . . . . . . . . 28

2 Applications of Induction and Recursion in Combinatorics and GraphTheory 312.1 Some Examples of Mathematical Induction . . . . . . . . . . . . . . . 31

2.1.1 Mathematical induction . . . . . . . . . . . . . . . . . . . . . . 312.1.2 Binomial Coefficients and the Binomial Theorem . . . . . . . . 332.1.3 Inductive definition . . . . . . . . . . . . . . . . . . . . . . . . 332.1.4 Proving the general product principle (Optional) . . . . . . . . 342.1.5 Double Induction and Ramsey Numbers . . . . . . . . . . . . 352.1.6 A bit of asymptotic combinatorics . . . . . . . . . . . . . . . . 36

2.2 Recurrence Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372.2.1 Examples of recurrence relations . . . . . . . . . . . . . . . . . 382.2.2 Arithmetic Series (optional) . . . . . . . . . . . . . . . . . . . . 392.2.3 First order linear recurrences . . . . . . . . . . . . . . . . . . . 402.2.4 Geometric Series . . . . . . . . . . . . . . . . . . . . . . . . . . 40

2.3 Graphs and Trees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

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2.3.1 Undirected graphs . . . . . . . . . . . . . . . . . . . . . . . . . 412.3.2 Walks and paths in graphs . . . . . . . . . . . . . . . . . . . . 432.3.3 Counting vertices, edges, and paths in trees . . . . . . . . . . . 432.3.4 Spanning trees . . . . . . . . . . . . . . . . . . . . . . . . . . . 452.3.5 Minimum cost spanning trees . . . . . . . . . . . . . . . . . . . 462.3.6 The deletion/contraction recurrence for spanning trees . . . . 472.3.7 Shortest paths in graphs . . . . . . . . . . . . . . . . . . . . . . 48

2.4 Supplementary Problems . . . . . . . . . . . . . . . . . . . . . . . . . 49

3 Distribution Problems 513.1 The idea of a distribution . . . . . . . . . . . . . . . . . . . . . . . . . . 51

3.1.1 The twentyfold way . . . . . . . . . . . . . . . . . . . . . . . . 513.1.2 Ordered functions . . . . . . . . . . . . . . . . . . . . . . . . . 543.1.3 Multisets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 563.1.4 Compositions of integers . . . . . . . . . . . . . . . . . . . . . 563.1.5 Broken permutations and Lah numbers . . . . . . . . . . . . . 57

3.2 Partitions and Stirling Numbers . . . . . . . . . . . . . . . . . . . . . . 583.2.1 Stirling Numbers of the second kind . . . . . . . . . . . . . . . 583.2.2 Stirling Numbers and onto functions . . . . . . . . . . . . . . . 603.2.3 Stirling Numbers and bases for polynomials . . . . . . . . . . 61

3.3 Partitions of Integers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 633.3.1 The number of partitions of k into n parts . . . . . . . . . . . . 633.3.2 Representations of partitions . . . . . . . . . . . . . . . . . . . 633.3.3 Ferrers and Young Diagrams and the conjugate of a partition 643.3.4 Partitions into distinct parts . . . . . . . . . . . . . . . . . . . . 69


4 Generating Functions 734.1 The Idea of Generating Functions . . . . . . . . . . . . . . . . . . . . . 73

4.1.1 Visualizing Counting with Pictures . . . . . . . . . . . . . . . 734.1.2 Picture functions . . . . . . . . . . . . . . . . . . . . . . . . . . 744.1.3 Generating functions . . . . . . . . . . . . . . . . . . . . . . . . 754.1.4 Power series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 774.1.5 Product principle for generating functions . . . . . . . . . . . 784.1.6 The extended binomial theorem and multisets . . . . . . . . . 79

4.2 Generating functions for integer partitions . . . . . . . . . . . . . . . . 814.3 Generating Functions and Recurrence Relations . . . . . . . . . . . . 85

4.3.1 How generating functions are relevant . . . . . . . . . . . . . . 854.3.2 Fibonacci numbers . . . . . . . . . . . . . . . . . . . . . . . . . 864.3.3 Second order linear recurrence relations . . . . . . . . . . . . . 864.3.4 Partial fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . 874.3.5 Catalan Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . 89


Contents xv

5 The Principle of Inclusion and Exclusion 935.1 The size of a union of sets . . . . . . . . . . . . . . . . . . . . . . . . . 93

5.1.1 Unions of two or three sets . . . . . . . . . . . . . . . . . . . . 935.1.2 Unions of an arbitrary number of sets . . . . . . . . . . . . . . 945.1.3 The Principle of Inclusion and Exclusion . . . . . . . . . . . . 95

5.2 Application of Inclusion and Exclusion . . . . . . . . . . . . . . . . . 975.2.1 Multisets with restricted numbers of elements . . . . . . . . . 975.2.2 The Menage Problem . . . . . . . . . . . . . . . . . . . . . . . . 975.2.3 Counting onto functions . . . . . . . . . . . . . . . . . . . . . . 975.2.4 The chromatic polynomial of a graph . . . . . . . . . . . . . . 98

5.3 Deletion-Contraction and the Chromatic Polynomial . . . . . . . . . . 995.4 Supplementary Problems . . . . . . . . . . . . . . . . . . . . . . . . . 101

6 Groups acting on sets 1036.1 Permutation Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

6.1.1 The rotations of a square . . . . . . . . . . . . . . . . . . . . . . 1036.1.2 Groups of Permutations . . . . . . . . . . . . . . . . . . . . . . 1046.1.3 The symmetric group . . . . . . . . . . . . . . . . . . . . . . . 1066.1.4 The dihedral group . . . . . . . . . . . . . . . . . . . . . . . . . 1076.1.5 Group tables (Optional) . . . . . . . . . . . . . . . . . . . . . . 1106.1.6 Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1116.1.7 The cycle structure of a permutation . . . . . . . . . . . . . . . 112

6.2 Groups Acting on Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . 1146.2.1 Groups acting on colorings of sets . . . . . . . . . . . . . . . . 1166.2.2 Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1196.2.3 The Cauchy-Frobenius-Burnside Theorem . . . . . . . . . . . 122

6.3 Pólya-Redfield Enumeration Theory . . . . . . . . . . . . . . . . . . . 1256.3.1 The Orbit-Fixed Point Theorem . . . . . . . . . . . . . . . . . . 1266.3.2 The Pólya-Redfield Theorem . . . . . . . . . . . . . . . . . . . 128


A Relations 133A.1 Relations as sets of Ordered Pairs . . . . . . . . . . . . . . . . . . . . . 133

A.1.1 The relation of a function . . . . . . . . . . . . . . . . . . . . . 133A.1.2 Directed graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . 135A.1.3 Digraphs of Functions . . . . . . . . . . . . . . . . . . . . . . . 136

A.2 Equivalence relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138

B Mathematical Induction 145B.1 The Principle of Mathematical Induction . . . . . . . . . . . . . . . . . 145

B.1.1 The ideas behind mathematical induction . . . . . . . . . . . . 145B.1.2 Mathematical induction . . . . . . . . . . . . . . . . . . . . . . 147B.1.3 Proving algebraic statements by induction . . . . . . . . . . . 148B.1.4 Strong Induction . . . . . . . . . . . . . . . . . . . . . . . . . . 149

xvi Contents

C Exponential Generating Functions 151C.1 Indicator Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151C.2 Exponential Generating Functions . . . . . . . . . . . . . . . . . . . . 152C.3 Applications to recurrences. . . . . . . . . . . . . . . . . . . . . . . . . 154

C.3.1 Using calculus with exponential generating functions . . . . . 155C.4 The Product Principle for EGFs . . . . . . . . . . . . . . . . . . . . . . 156C.5 The Exponential Formula . . . . . . . . . . . . . . . . . . . . . . . . . 162C.6 Supplementary Problems . . . . . . . . . . . . . . . . . . . . . . . . . 166

D Hints to Selected Problems 169

E GNU Free Documentation License 193

Index 199

Chapter 1

What is Combinatorics?

Combinatorial mathematics arises from studying how we can combine objects intoarrangements. For example, we might be combining sports teams into a tourna-ment, samples of tires into plans to mount them on cars for testing, students intoclasses to compare approaches to teaching a subject, or members of a tennis clubinto pairs to play tennis. There are many questions one can ask about such ar-rangements of objects. Here we will focus on questions about how many ways wemay combine the objects into arrangements of the desired type. These are calledcounting problems. Sometimes, though, combinatorial mathematicians ask if anarrangement is possible (if we have ten baseball teams, and each team has to playeach other team once, can we schedule all the games if we only have the fields avail-able at enough times for forty games?). Sometimes they ask if all the arrangementswe might be able to make have a certain desirable property (Do all ways of testing 5brands of tires on 5 different cars [with certain additional properties] compare eachbrand with each other brand on at least one common car?). Problems of these sortscome up throughout physics, biology, computer science, statistics, and many othersubjects. However, to demonstrate all these relationships, we would have to takedetours into all these subjects. While we will give some important applications, wewill usually phrase our discussions around everyday experience and mathematicalexperience so that the student does not have to learn a new context before learningmathematics in context!

1.1 About These NotesThese notes are based on the philosophy that you learn the most about a subjectwhen you are figuring it out directly for yourself, and learn the least when you aretrying to figure out what someone else is saying about it. On the other hand, thereis a subject called combinatorial mathematics, and that is what we are going to bestudying, so we will have to tell you some basic facts. What we are going to try todo is to give you a chance to discover many of the interesting examples that usuallyappear as textbook examples and discover the principles that appear as textbooktheorems. Your main activity will be solving problems designed to lead you todiscover the basic principles of combinatorial mathematics. Some of the problemslead you through a new idea, some give you a chance to describe what you have

1

2 1. What is Combinatorics?

learned in a sequence of problems, and some are quite challenging. When you finda problem challenging, don’t give up on it, but don’t let it stop you from going onwith other problems. Frequently you will find an idea in a later problem that youcan take back to the one you skipped over or only partly finished in order to finishit off. With that in mind, let’s get started. In the problems that follow, you willsee some problems marked on the left with various symbols. The preface gives afull explanation of these symbols and discusses in greater detail why the book isorganized as it is! Table 1.1.1, which is repeated from the preface, summarizes themeaning of the symbols.

• essential◦ motivational material+ summary⇒ especially interesting∗ difficult· essential for this section or the next

Table 1.1.1: The meaning of the symbols to the left of problem numbers.

1.2 Basic Counting Principles

Problem 1.◦ Five schools are going to send their baseball teams to a tourna-ment, in which each team must play each other team exactly once. Howmany games are required? (h)

Problem 2.• Now some number n of schools are going to send their baseballteams to a tournament, and each team must play each other team exactlyonce. Let us think of the teams as numbered 1 through n.

(a) How many games does team 1 have to play in?

(b) How many games, other than the one with team 1, does team twohave to play in?

(c) How many games, other than those with the first i − 1 teams, doesteam i have to play in?

(d) In terms of your answers to the previous parts of this problem, whatis the total number of games that must be played?

Problem 3.• One of the schools sending its team to the tournament has tosend its players from some distance, and so it is making sandwiches for

1.2. Basic Counting Principles 3

team members to eat along the way. There are three choices for the kind ofbread and five choices for the kind of filling. How many different kinds ofsandwiches are available? (h)

Problem 4.+ An ordered pair (a , b) consists of two things we call a and b. Wesay a is the first member of the pair and b is the second member of the pair.If M is an m element set and N is an n-element set, how many ordered pairsare there whose first member is in M and whose second member is in N?Does this problem have anything to do with any of the previous problems?

Problem 5.◦ Since a sandwich by itself is pretty boring, students from theschool in Problem 3 are offered a choice of a drink (from among five differentkinds), a sandwich, and a fruit (from among four different kinds). In howmany ways may a student make a choice of the three items now?

Problem 6.• The coach of the team in Problem 3 knows of an ice cream parloralong the way where she plans to stop to buy each team member a tripledecker cone. There are 12 different flavors of ice cream, and triple deckercones are made in homemade waffle cones. Having chocolate ice cream asthe bottom scoop is different from having chocolate ice cream as the topscoop. How many possible ice cream cones are going to be available to theteam members? How many cones with three different kinds of ice creamwill be available? (h)

Problem 7.• The idea of a function is ubiquitous in mathematics. A functionf from a set S to a set T is a relationship between the two sets that associatesexactly one member f (x) of T with each element x in S. We will comeback to the ideas of functions and relationships in more detail and fromdifferent points of view from time to time. However, the quick review aboveshould probably let you answer these questions. If you have difficulty withthem, it would be a good idea to go now to Appendix A and work throughSection A.1 which covers this definition in more detail. You might also wantto study Section A.1.3 to learn to visualize the properties of functions. Wewill take up the topic of this section later in this chapter as well, but in lessdetail than is in the appendix.

(a) Using f , g, . . . , to stand for the various functions, write down allthe different functions you can from the set {1, 2} to the set {a , b}.


For example, you might start with f (1) = a, f (2) = b. How manyfunctions are there from the set {1, 2} to the set {a , b}? (h)

(b) How many functions are there from the three element set {1, 2, 3} tothe two element set {a , b}? (h)

(c) How many functions are there from the two element set {a , b} to thethree element set {1, 2, 3}? (h)

(d) How many functions are there from a three element set to a 12 elementset?

(e) The function f is called one-to-one or an injection if whenever x isdifferent from y, f (x) is different from f (y). How many one-to-onefunctions are there from a three element set to a 12 element set?

(f) Explain the relationship between this problem and Problem 6.

Problem 8.• A group of hungry team members in Problem 6 notices it wouldbe cheaper to buy three pints of ice cream for them to split than to buy atriple decker cone for each of them, and that way they would get more icecream. They ask their coach if they can buy three pints of ice cream.

(a) In how many ways can they choose three pints of different flavors outof the 12 flavors? (h)

(b) In how many ways may they choose three pints if the flavors don’thave to be different? (h)

Problem 9.• Two sets are said to be disjoint if they have no elements incommon. For example, {1, 3, 12} and {6, 4, 8, 2} are disjoint, but {1, 3, 12}and {3, 5, 7} are not. Three or more sets are said to be mutually disjoint ifno two of them have any elements in common. What can you say about thesize of the union of a finite number of finite (mutually) disjoint sets? Doesthis have anything to do with any of the previous problems?

Problem 10.• Disjoint subsets are defined in Problem 9. What can you sayabout the size of the union of m (mutually) disjoint sets, each of size n?Does this have anything to do with any of the previous problems?


1.2.1 The sum and product principlesThese problems contain among them the kernels of many of the fundamentalideas of combinatorics. For example, with luck, you just stated the sum principle(illustrated in Figure 1.2.1), and product principle (illustrated in Figure 1.2.2) inProblems 9 and Problem 10. These two counting principles are the basis on whichwe will develop many other counting principles.

Figure 1.2.1: The union of these two disjoint sets has size 17.

Figure 1.2.2: The union of four disjoint sets of size five.

You may have noticed some standard mathematical words and phrases suchas set, ordered pair, function and so on creeping into the problems. One of ourgoals in these notes is to show how most counting problems can be recognized ascounting all or some of the elements of a set of standard mathematical objects. Forexample Problem 4 is meant to suggest that the question we asked in Problem 3was really a problem of counting all the ordered pairs consisting of a bread choiceand a filling choice. We use A×B to stand for the set of all ordered pairs whose firstelement is in A and whose second element is in B and we call A × B the Cartesianproduct of A and B, so you can think of Problem 4 as asking you for the size of theCartesian product of M and N , that is, asking you to count the number of elementsof this Cartesian product.

When a set S is a union of disjoint sets B1 , B2 , . . . , Bm we say that the setsB1 , B2 , . . . , Bm are a partition of the set S. Thus a partition of S is a (special kindof) set of sets. So that we don’t find ourselves getting confused between the set S


and the sets Bi into which we have divided it, we often call the sets B1 , B2 , . . . , Bmthe blocks of the partition. In this language, the sum principle says that

if we have a partition of a set S, then the size of S is the sum of the sizesof the blocks of the partition.

The product principle says that

if we have a partition of a set S into m blocks, each of size n, then S hassize mn.

You’ll notice that in our formal statement of the sum and product pinciple wetalked about a partition of a finite set. We could modify our language a bit to coverinfinite sizes, but whenever we talk about sizes of sets in what follows, we will beworking with finite sets. So as to avoid possible complications in the future, letus agree that when we refer to the size of a set, we are implicitly assuming the setis finite. There is another version of the product principle that applies directly inproblems like Problem 5 and Problem 6, where we were not just taking a union ofm disjoint sets of size n, but rather m disjoint sets of size n, each of which was aunion of m′ disjoint sets of size n′. This is an inconvenient way to have to thinkabout a counting problem, so we may rephrase the product principle in terms of asequence of decisions:

Problem 11.• If we make a sequence of m choices for which

• there are k1 possible first choices, and

• for each way of making the first i−1 choices, there are ki ways to makethe ith choice,

then in how many ways may we make our sequence of choices? (You neednot prove your answer correct at this time.)

The counting principle you gave in Problem 11 is called the general productprinciple. We will outline a proof of the general product pinciple from the originalproduct principle in Problem 80. Until then, let us simply accept it as anothercounting principle. For now, notice how much easier it makes it to explain why wemultiplied the things we did in Problem 5 and Problem 6.

Problem 12.⇒ A tennis club has 2n members. We want to pair up the mem-bers by twos for singles matches.

(a) In how many ways may we pair up all the members of the club? (Hint:consider the cases of 2, 4, and 6 members.) (h)

(b) Suppose that in addition to specifying who plays whom, for eachpairing we say who serves first. Now in how many ways may wespecify our pairs?


Problem 13.+ Let us now return to Problem 7 and justify—or perhaps fin-ish—our answer to the question about the number of functions from athree-element set to a 12-element set.

(a) How can you justify your answer in Problem 7 to the question “Howmany functions are there from a three element set (say [3] = {1, 2, 3})to a twelve element set (say [12])?”

(b) Based on the examples you’ve seen so far, make a conjecture abouthow many functions there are from the set

[m] = {1, 2, 3, . . . ,m}

to [n] = {1, 2, 3, . . . , n} and prove it.

(c) A common notation for the set of all functions from a set M to a set Nis NM . Why is this a good notation?

Problem 14.+ Now suppose we are thinking about a set S of functions f from[m] to some set X. (For example, in Problem 6 we were thinking of the set offunctions from the three possible places for scoops in an ice-cream cone to12 flavors of ice cream.) Suppose there are k1 choices for f (1). (In Problem 6,k1 was 12, because there were 12 ways to choose the first scoop.) Supposethat for each choice of f (1) there are k2 choices for f (2). (For example, inProblem 6 k2 was 12 if the second flavor could be the same as the first, butk2 was 11 if the flavors had to be different.) In general, suppose that for eachchoice of f (1), f (2), . . . f (i − 1), there are ki choices for f (i). (For example,in Problem 6, if the flavors have to be different, then for each choice of f (1)and f (2), there are 10 choices for f (3).)What we have assumed so far about the functions in S may be summarizedas

• There are k1 choices for f (1).

• For each choice of f (1), f (2), . . . , f (i − 1), there are ki choices for f (i).

How many functions are in the set S? Is there any practical differencebetween the result of this problem and the general product principle?

The point of Problem 14 is that the general product principle can be statedinformally, as we did originally, or as a statement about counting sets of standardconcrete mathematical objects, namely functions.

Problem 15.⇒ A roller coaster car has n rows of seats, each of which has roomfor two people. If n men and n women get into the car with a man and awoman in each row, in how many ways may they choose their seats? (h)


Problem 16.+ How does the general product principle apply to Problem 6?

Problem 17.• In how many ways can we pass out k distinct pieces of fruit ton children (with no restriction on how many pieces of fruit a child may get)?

Problem 18.• How many subsets does a set S with n elements have? (h)

Problem 19.◦ Assuming k ≤ n, in how many ways can we pass out k distinctpieces of fruit to n children if each child may get at most one? What is thenumber if k > n? Assume for both questions that we pass out all the fruit. (h)

Problem 20.• Another name for a list, in a specific order, of k distinct thingschosen from a set S is a k-element permutation of S. We can also think ofa k-element permutation of S as a one-to-one function (or, in other words,injection) from [k] = {1, 2, . . . , k} to S. How many k-element permutationsdoes an n-element set have? (For this problem it is natural to assume k ≤ n.However the question makes sense even if k > n. What is the number ofk-element permutations of an n-element set if k > n? (h)

There are a number of different notations for the number of k-element permu-tations of an n-element set. The one we shall use was introduced by Don Knuth;namely nk , read “n to the k falling” or “n to the k down”. In Problem 20 you mayhave shown that

nk = n(n − 1) · · · (n − k + 1) =k∏

i=1

(n − i + 1). (1.1)

It is standard to call nk the k-th falling factorial power of n, which explainswhy we use exponential notation. Of course we call it a factorial power since nn =n(n − 1) · · · 1 which we call n-factorial and denote by n!. If you are unfamiliar withthe Π notation, or product notation we introduced for products in Equation (1.1),it works just like the Σ notation works for summations.

Problem 21.• Express nk as a quotient of factorials.

Problem 22.⇒ How should we define n0?


1.2.2 Functions and directed graphsAs another example how standard mathematical language relates to counting prob-lems, Problem 7 explicitly asked you to relate the idea of counting functions to thequestion of Problem 6. You have probably learned in algebra or calculus how todraw graphs in the Cartesian plane of functions from a set of numbers to a set ofnumbers. You may recall how we can determine whether a graph is a graph of afunction by examining whether each vertical straight line crosses the graph at mostone time. You might also recall how we can determine whether such a function isone-to-one by examining whether each horizontal straight line crosses the graph atmost one time. The functions we deal with will often involve objects which are notnumbers, and will often be functions from one finite set to another. Thus graphsin the cartesian plane will often not be available to us for visualizing functions.

However, there is another kind of graph called a directed graph or digraphthat is especially useful when dealing with functions between finite sets. Wetake up this topic in more detail in Appendix A, particularly Subsection A.1.2 andSubsection A.1.3. In Figure 1.2.3 we show several examples of digraphs of functions.

(e) The function from {0, 1, 2, 3, 4, 5} to

{0, 1, 2, 3, 4, 5} given by f(x) = x + 2 mod 6

Figure 1.2.3: What is a digraph of a function?


If we have a function f from a set S to a set T, we draw a line of dots or circles,called vertices to represent the elements of S and another (usually parallel) line ofcircles or dots to represent the elements of T. We then draw an arrow from thecircle for x to the circle for y if f (x) = y. Sometimes, as in part (e) of the figure, if wehave a function from a set S to itself, we draw only one set of vertices representingthe elements of S, in which case we can have arrows both entering and leaving agiven vertex. As you see, the digraph can be more enlightening in this case if weexperiment with the function to find a nice placement of the vertices rather thanputting them in a row.

Notice that there is a simple test for whether a digraph whose vertices representthe elements of the sets S and T is the digraph of a function from S to T. Theremust be one and only one arrow leaving each vertex of the digraph representing anelement of S. The fact that there is one arrow means that f (x) is defined for each xin S. The fact that there is only one arrow means that each x in S is related to exactlyone element of T. (Note that these remarks hold as well if we have a function fromS to S and draw only one set of vertices representing the elements of S.) For furtherdiscussion of functions and digraphs see Sections A.1.1 and Subsection A.1.2 ofAppendix A.

Problem 23.◦ Draw the digraph of the function from the set {Alice, Bob,Dawn, Bill} to the set {A, B, C, D, E} given by

f (X) = the first letter of the name X.

Problem 24.• A function f : S → T is called an onto function or surjectionif each element of T is f (x) for some x ∈ S. Choose a set S and a set T sothat you can draw the digraph of a function from S to T that is one-to-onebut not onto, and draw the digraph of such a function.

Problem 25.◦ Choose a set S and a set T so that you can draw the digraph ofa function from S to T that is onto but not one-to-one, and draw the digraphof such a function.

Problem 26.• Digraphs of functions help us visualize the ideas of one-to-onefunctions and onto functions.

(a) What does the digraph of a one-to-one function (injection) from a finiteset X to a finite set Y look like? (Look for a test somewhat similar to theone we described for when a digraph is the digraph of a function.) (h)

(b) What does the digraph of an onto function look like?


(c) What does the digraph of a one-to-one and onto function from a finiteset S to a set T look like?

Problem 27.• The word permutation is actually used in two different ways inmathematics. A permutation of a set S is one-to-one from S onto S. Howmany permutations does an n-element set have?

Notice that there is a great deal of consistency between the use of the wordpermutation in Problem 27 and the use in Problem 20. If we have some waya1 , a2 , . . . , an of listing our set, then any other list b1 , b2 , . . . , bn gives us the permu-tation of S whose rule is f (ai) = bi , and any permutation of S, say the one given byg(ai) = ci gives us a list c1 , c2 , . . . , cn of S. Thus there is really very little differencebetween the idea of a permutation of S and an n-element permutation of S whenn is the size of S.

1.2.3 The bĳection principleAnother name for a one-to-one and onto function is bĳection. The digraphs marked(a), (b), and (e) in Figure 1.2.3 are digraphs of bĳections. The description in Prob-lem 26.c of the digraph of a bĳection from X to Y illustrates one of the fundamentalprinciples of combinatorial mathematics, the bĳection principle:

Two sets have the same size if and only if there is a bĳection betweenthem.

It is surprising how this innocent sounding principle guides us into finding insightinto some otherwise very complicated proofs.

1.2.4 Counting subsets of a set

Problem 28. The binary representation of a number m is a list, or string,a1a2 . . . ak of zeros and ones such that m = a12k−1 + a22k−2 + · · · + ak20.Describe a bĳection between the binary representations of the integers be-tween 0 and 2n − 1 and the subsets of an n-element set. What does this tellyou about the number of subsets of an n-element set? (h)

Notice that the first question in Problem 8 asked you for the number of waysto choose a three element subset from a 12 element subset. You may have seen anotation like (n

k ), C(n , k), or nCk which stands for the number of ways to choose a k-element subset from an n-element set. The number (n

k ) is read as “n choose k” andis called a binomial coefficient for reasons we will see later on. Another frequentlyused way to read the binomial coefficient notation is “the number of combinationsof n things taken k at a time." You are going to be asked to construct two bĳectionsthat relate to these numbers and figure out what famous formula they prove. We


are going to think about subsets of the n-element set [n] = {1, 2, 3, . . . , n}. As anexample, the set of two-element subsets of [4] is

{{1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}}.

This example tells us that (42 ) = 6.

Problem 29.• Let C be the set of k-element subsets of [n] that contain thenumber n, and let D be the set of k-element subsets of [n] that don’t containn.

(a) Let C′ be the set of (k−1)-element subsets of [n−1]. Describe a bĳectionfrom C to C′. (A verbal description is fine.)

(b) Let D′ be the set of k-element subsets of [n − 1] = {1, 2, . . . n − 1}.Describe a bĳection from D to D′. (A verbal description is fine.)

(c) Based on the two previous parts, express the sizes of C and D in termsof binomial coefficients involving n − 1 instead of n.

(d) Apply the sum principle to C and D and obtain a formula that ex-presses (n

k ) in terms of two binomial coefficients involving n − 1. Youhave just derived the Pascal Equation that is the basis for the famousPascal’s Triangle.

1.2.5 Pascal’s TriangleThe Pascal Equation that you derived in Problem 29 gives us the triangle in Fig-ure 1.2.4. This figure has the number of k-element subsets of an n-element set asthe kth number over in the nth row (we call the top row the zeroth row and thebeginning entry of a row the zeroth number over). You’ll see that your formuladoesn’t say anything about (n

k ) if k = 0 or k = n, but otherwise it says that eachentry is the sum of the two that are above it and just to the left or right.

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

1 7 21 35 35 21 7 1

Figure 1.2.4: Pascal’s Triangle


Problem 30. Just for practice, what is the next row of Pascal’s triangle?

Problem 31.⇒ Without writing out the rows completely, write out enough ofPascal’s triangle to get a numerical answer for the first question in Prob-lem 8. (h)

It is less common to see Pascal’s triangle as a right triangle, but it actuallymakes your formula easier to interpret. In Pascal’s Right Triangle, the element inrow n and column k (with the convention that the first row is row zero and thefirst column is column zero) is (n

k ). In this case your formula says each entry in arow is the sum of the one above and the one above and to the left, except for theleftmost and rightmost entries of a row, for which that doesn’t make sense. Sincethe leftmost entry is (n

0 ) and the rightmost entry is (nn ), these entries are both one

(to see why, ask yourself how many 0-element subsets and how many n-elementsubsets an n-element set has), and your formula then tells how to fill in the rest ofthe table.

k = 0 1 2 3 4 5 6 7n = 0 1

1 1 12 1 2 13 1 3 3 14 1 4 6 4 15 1 5 10 10 5 16 1 6 15 20 15 6 17 1 7 21 35 35 21 7 1

Table 1.2.5: Pascal’s Right Triangle

Seeing this right triangle leads us to ask whether there is some natural wayto extend the right triangle to a rectangle. If we did have a rectangular table ofbinomial coefficients, counting the first row as row zero (i.e., n = 0) and the firstcolumn as column zero (i.e., k = 0), the entries we don’t yet have are values of (n

k )for k > n. But how many k-element subsets does an n-element set have if k > n?The answer, of course, is zero, so all the other entries we would fill in would bezero, giving us the rectangular array in Figure 1.2.6. It is straightforward to checkthat Pascal’s equation now works for all the entries in the rectangle that have anentry above them and an entry above and to the left.


k = 0 1 2 3 4 5 6 7n = 0 1 0 0 0 0 0 0 0

1 1 1 0 0 0 0 0 02 1 2 1 0 0 0 0 03 1 3 3 1 0 0 0 04 1 4 6 4 1 0 0 05 1 5 10 10 5 1 0 06 1 6 15 20 15 6 1 07 1 7 21 35 35 21 7 1

Table 1.2.6: Pascal’s Rectangle

Problem 32.⇒ Because our definition told us that (nk ) is 0 when k > n, we got

a rectangular table of numbers that satisfies the Pascal Equation.

(a) Is there any other way to define (nk ) when k > n in order to get a

rectangular table that agrees with Pascal’s Right Triangle for k ≤ nand satisfies the Pascal Equation? (h)

(b) Suppose we want to extend Pascal’s Rectangle to the left and define( n−k ) for n ≥ 0 and k > 0 so that −k < 0. What should we put into row

n and column −k of Pascal’s Rectangle in order for the Pascal Equationto hold true? (h)

(c) What should we put into row −n and column k or column −k in orderfor the Pascal Equation to continue to hold? Do we have any freedomof choice? (h)

Problem 33. There is yet another bĳection that lets us prove that a set ofsize n has 2n subsets. Namely, for each subset S of [n] = {1, 2, . . . , n}, definea function (traditionally denoted by χS) as follows.a

χS(i) =

{1 if i ∈ S0 if i ! S

The function χS is called the characteristic function of S. Notice that thecharacteristic function is a function from [n] to {0, 1}.

(a) For practice, consider the function χ{1,3} for the subset {1, 3} of the set{1, 2, 3, 4}. What are

(i) χ{1,3}(1)?(ii) χ{1,3}(2)?

(iii) χ{1,3}(3)?(iv) χ{1,3}(4)?


(b) We define a function f from the set of subsets of [n] = {1, 2, . . . , n} tothe set of functions from [n] to {0, 1} by f (S) = χS. Explain why f isa bĳection.

(c) Why does the fact that f is a bĳection prove that [n] has 2n subsets?aThe symbol χ is the Greek letter chi that is pronounced Ki, with the i sounding like “eye.”

In Problems 18, Problem 28, and Problem 33 you gave three proofs of thefollowing theorem.

Theorem 1.2.7. The number of subsets of an n-element set is 2n .

The proofs in Problem 28 and Problem 33 use essentially the same bĳection,but they interpret sequences of zeros and ones differently, and so end up beingdifferent proofs. We will give yet another proof, using bĳections similar to thosewe used in proving the Pascal Equation, at the beginning of Chapter 2.

1.2.6 The quotient principle

Problem 34.• As we noted in Problem 29, the first question in Problem 8asked us for the number of three-element subsets of a twelve-element set.We were able to use the Pascal Equation to get a numerical answer to thatquestion. Had we had twenty or thirty flavors of ice cream to choose from,using the Pascal Equation to get our answer would have entailed a goodbit more work. We have seen how the general product principle gives usan answer to Problem 6. Thus we might think that the number of waysto choose a three element set from 12 elements is the number of ways tochoose the first element times the number of ways to choose the secondelement times the number of ways to choose the third element, which is12 · 11 · 10 = 1320. However, our result in Problem 29 shows that this iswrong.

(a) What is it that is different between the number of ways to stack icecream in a triple decker cone with three different flavors of ice creamand the number of ways to simply choose three different flavors of icecream?

(b) In particular, how many different triple decker cones use the samethree flavors? (Of course any three distinct flavors could substitute forvanilla, chocolate and strawberry without changing the answer.)

(c) Using your answer from part b, compute the number of ways to choosethree different flavors of ice cream (out of twelve flavors) from thenumber of ways to choose a triple decker cone with three differentflavors (out of twelve flavors).


Problem 35.• Based on what you observed in Problem 34.c, how many k-element subsets does an n-element set have?

Problem 36.⇒ The formula you proved in Problem 35 is symmetric in k andn− k; that is, it gives the same number for (n

k ) as it gives for ( nn−k ). Whenever

two quantities are counted by the same formula it is good for our insightto find a bĳection that demonstrates the two sets being counted have thesame size. In fact this is a guiding principle of research in combinatorialmathematics. Find a bĳection that proves that (n

k ) equals ( nn−k ). (h)

Problem 37.• In how many ways can we pass out k (identical) ping-pongballs to n children if each child may get at most one? (h)

Problem 38.• In how many ways may n people sit around a round table?(Assume that when people are sitting around a round table, all that reallymatters is who is to each person’s right. For example, if we can get onearrangement of people around the table from another by having everyoneget up and move to the right one place and sit back down, we get anequivalent arrangement of people. Notice that you can get a list from aseating arrangement by marking a place at the table, and then listing thepeople at the table, starting at that place and moving around to the right.)There are at least two different ways of doing this problem. Try to find themboth. (h)

We are now going to analyze the result of Problem 35 in more detail in order totease out another counting principle that we can use in a wide variety of situations.

abc acb bac bca cab cbaabd adb bad bda dab dbaabe aeb bae bea eab ebaacd adc cad cda dac dcaace aec cae cea eac ecaade aed dae dea ead edabcd bdc cbd cdb dbc dcbbce bec cbe ceb ebc ecbbde bed dbe deb ebd edbcde ced dce dec ecd edc

Table 1.2.8: The 3-element permutations of {a , b , c , d , e} organized by which 3-element set they permute.


In Table 1.2.8 we list all three-element permutations from the 5-element set{a , b , c , d , e}. Each row consists of all 3-element permutations of some subsetof {a , b , c , d , e}. Because a given k-element subset can be listed as a k-elementpermutation in k! ways, there are 3! = 6 permutations in each row. Because each3-element permutation appears exactly once in the table, each row is a block of apartition of the set of 3-element permutations of {a , b , c , d , e}. Each block has sizesix. Each block consists of all 3-element permutations of some three-element subsetof {a , b , c , d , e}. Since there are ten rows, we see that there are ten 3-element subsetsof {a , b , c , d , e}. An alternate way to see this is to observe that we partitioned the setof all 60 three-element permutations of {a , b , c , d , e} into some number q of blocks,each of size six. Thus by the product principle, q · 6 = 60, so q = 10.

Problem 39.• Rather than restricting ourselves to n = 5 and k = 3, we canpartition the set of all k-element permutations of S up into blocks. We do soby letting BK be the set (block) of all k-element permutations of K for each k-element subset K of S. Thus as in our preceding example, each block consistsof all permutations of some subset K of our n-element set. For example,the permutations of {a , b , c} are listed in the first row of Table 1.2.8. In facteach row of that table is a block. The questions that follow are about thecorresponding partition of the set of k-element permutations of S, where Sand k are arbitrary.

(a) How many permutations are there in a block? (h)

(b) Since S has n elements, what does problem 20 tell you about the totalnumber of k-element permutations of S?

(c) Describe a bĳection between the set of blocks of the partition and theset of k-element subsets of S. (h)

(d) What formula does this give you for the number (nk ) of k-element

subsets of an n-element set? (h)

Problem 40.⇒ A basketball team has 12 players. However, only five playersplay at any given time during a game.

(a) In how may ways may the coach choose the five players?

(b) To be more realistic, the five players playing a game normally consistof two guards, two forwards, and one center. If there are five guards,four forwards, and three centers on the team, in how many ways canthe coach choose two guards, two forwards, and one center? (h)

(c) What if one of the centers is equally skilled at playing forward? (h)


Problem 41.• In Problem 38, describe a way to partition the n-element per-mutations of the n people into blocks so that there is a bĳection betweenthe set of blocks of the partition and the set of arrangements of the n peoplearound a round table. What method of solution for Problem 38 does thiscorrespond to?

Problem 42.• In Problems 39.d and 41, you have been using the productprinciple in a new way. One of the ways in which we previously stated theproduct principle was “If we partition a set into m blocks each of size n,then the set has size m · n.” In Problems 39.d and 41 we knew the size p of aset P of permutations of a set, and we knew we had partitioned P into someunknown number of blocks, each of a certain known size r. If we let q standfor the number of blocks, what does the product principle tell us about p, q,and r? What do we get when we solve for q?

The formula you found in the Problem 42 is so useful that we are going to singleit out as another principle. The quotient principle says:

If we partition a set P into q blocks, each of size r, then q = p/r.

The quotient principle is really just a restatement of the product principle, butthinking about it as a principle in its own right often leads us to find solutions toproblems. Notice that it does not always give us a formula for the number of blocksof a partition; it only works when all the blocks have the same size. In Chapter 6,we develop a way to solve problems with different block sizes in cases where thereis a good deal of symmetry in the problem. (The roundness of the table was asymmetry in the problem of people at a table; the fact that we can order the sets inany order is the symmetry in the problem of counting k-element subsets.)

In Section A.2 of Appendix A we introduce the idea of an equivalence relation,see what equivalence relations have to do with partitions, and discuss the quotientprinciple from that point of view. While that appendix is not required for what weare doing here, if you want a more thorough discussion of the quotient principle,this would be a good time to work through that appendix.

Problem 43. In how many ways may we string n distinct beads on a necklacewithout a clasp? (Perhaps we make the necklace by stringing the beads on astring, and then carefully gluing the two ends of the string together so thatthe joint can’t be seen. Assume someone can pick up the necklace, move itaround in space and put it back down, giving an apparently different wayof stringing the beads that is equivalent to the first.) (h)


Problem 44.⇒ We first gave this problem as Problem 12.a. Now we haveseveral ways to approach the problem. A tennis club has 2n members. Wewant to pair up the members by twos for singles matches.

(a) In how many ways may we pair up all the members of the club? Giveat least two solutions different from the one you gave in Problem 12.a.(You may not have done Problem 12.a. In that case, see if you can findthree solutions.) (h)

(b) Suppose that in addition to specifying who plays whom, for eachpairing we say who serves first. Now in how many ways may wespecify our pairs? Try to find as many solutions as you can. (h)

Problem 45.• (This becomes especially relevant in Chapter 6, though itmakes an important point here.) In how many ways may we attach twoidentical red beads and two identical blue beads to the corners of a square(with one bead per corner) free to move around in (three-dimensional)space? (h)

Problem 46.⇒ While the formula you proved in Problems 35 and 39.d is veryuseful, it doesn’t give us a sense of how big the binomial coefficients are.We can get a very rough idea, for example, of the size of (2n

n ) by recognizingthat we can write (2n)n/n! as 2n

n · 2n−1n−1 · · · n+1

1 , and each quotient is at least2, so the product is at least 2n . If this were an accurate estimate, it wouldmean the fraction of n-element subsets of a 2n-element set would be about2n/22n = 1/2n , which becomes very small as n becomes large. However itis pretty clear the approximation will not be a very good one, because someof the terms in that product are much larger than 2. In fact, if (2n

k ) were thesame for every k, then each would be the fraction 1

2n+1 of 22n . This is muchlarger than the fraction 1

2n . But our intuition suggets that (2nn ) is much larger

than (2n1 ) and is likely larger than ( 2n

n−1 ) so we can be sure our approximationis a bad one. For estimates like this, James Stirling developed a formula toapproximate n! when n is large, namely n! is about

(√2πn

)nn/en . In fact

the ratio of n! to this expression approaches 1 as n becomes infinite.a Wewrite this as

n! ∼√2πn

nn

en .

We read this notation as n! is asymptotic to√2πn nn

en . Use Stirling’s for-mula to show that the fraction of subsets of size n in an 2n-element set isapproximately 1/

√πn. This is a much bigger fraction than 1

2n !


aProving this takes more of a detour than is advisable here; however there is an elementaryproof which you can work through in the problems of the end of Section 1 of Chapter 1 ofIntroductory Combinatorics by Kenneth P. Bogart, Harcourt Academic Press, (2000).

1.3 Some Applications of the Basic Principles1.3.1 Lattice paths and Catalan Numbers

Problem 47.◦ In a part of a city, all streets run either north-south or east-west,and there are no dead ends. Suppose we are standing on a street corner. Inhow many ways may we walk to a corner that is four blocks north and sixblocks east, using as few blocks as possible? (h)

Problem 48.· Problem 47 has a geometric interpretation in a coordinateplane. A lattice path in the plane is a “curve” made up of line segmentsthat either go from a point (i , j) to the point (i + 1, j) or from a point (i , j)to the point (i , j + 1), where i and j are integers. (Thus lattice paths alwaysmove either up or to the right.) The length of the path is the number of suchline segments.

(a) What is the length of a lattice path from (0, 0) to (m , n)?

(b) How many such lattice paths of that length are there? (h)

(c) How many lattice paths are there from (i , j) to (m , n), assuming i, j,m, and n are integers? (h)

Problem 49.· Another kind of geometric path in the plane is a diagonallattice path. Such a path is a path made up of line segments that go froma point (i , j) to (i + 1, j + 1) (this is often called an upstep) or (i + 1, j − 1)(this is often called a downstep), again where i and j are integers. (Thusdiagonal lattice paths always move towards the right but may move up ordown.)

(a) Describe which points are connected to (0, 0) by diagonal latticepaths. (h)

(b) What is the length of a diagonal lattice path from (0, 0) to (m , n)?

(c) Assuming that (m , n) is such a point, how many diagonal lattice pathsare there from (0, 0) to (m , n)? (h)

1.3. Some Applications of the Basic Principles 21

Problem 50.◦ A school play requires a ten dollar donation per person; thedonation goes into the student activity fund. Assume that each person whocomes to the play pays with a ten dollar bill or a twenty dollar bill. Theteacher who is collecting the money forgot to get change before the event.If there are always at least as many people who have paid with a ten asa twenty as they arrive the teacher won’t have to give anyone an IOU forchange. Suppose 2n people come to the play, and exactly half of them paywith ten dollar bills.

(a) Describe a bĳection between the set of sequences of tens and twentiespeople give the teacher and the set of lattice paths from (0, 0) to (n , n).

(b) Describe a bĳection between the set of sequences of tens and twentiesthat people give the teacher and the set of diagonal lattice paths from(0, 0) and (2n , 0).

(c) In each case, what is the geometric interpretation of a sequence thatdoes not require the teacher to give any IOUs? (h)

Problem 51.⇒ · Notice that a lattice path from (0, 0) to (n , n) stays inside (oron the edges of) the square whose sides are the x-axis, the y-axis, the linex = n and the line y = n. In this problem we will compute the numberof lattice paths from (0,0) to (n , n) that stay inside (or on the edges of) thetriangle whose sides are the x-axis, the line x = n and the line y = x. Forexample, in Figure 1.3.1 we show the grid of points with integer coordinatesfor the triangle whose sides are the x-axis, the line x = 4 and the line y = x.

1

1

2

5

14

Figure 1.3.1: The lattice paths from (0, 0) to (i , i) for i = 0, 1, 2, 3, 4. Thenumber of paths to the point (i , i) is shown just above that point.

(a) Explain why the number of lattice paths from (0, 0) to (n , n) that gooutside the triangle is the number of lattice paths from (0, 0) to (n , n)that either touch or cross the line y = x + 1.

(b) Find a bĳection between lattice paths from (0, 0) to (n , n) that touch(or cross) the line y = x + 1 and lattice paths from (−1, 1) to (n , n). (h)


(c) Find a formula for the number of lattice paths from (0, 0) to (n , n) thatdo not go above the line y = x. The number of such paths is called aCatalan Number and is usually denoted by Cn . (h)

Problem 52.⇒ Your formula for the Catalan Number can be expressed as abinomial coefficient divided by an integer. Whenever we have a formulathat calls for division by an integer, an ideal combinatorial explanation ofthe formula is one that uses the quotient principle. The purpose of thisproblem is to find such an explanation using diagonal lattice paths.a Adiagonal lattice path that never goes below the y-coordinate of its first pointis called a Dyck Path. We will call a Dyck Path from (0, 0) to (2n , 0) aCatalan Path of length 2n. Thus the number of Catalan Paths of length 2nis the Catalan Number Cn .

(a) If a Dyck Path has n steps (each an upstep or downstep), why do thefirst k steps form a Dyck Path for each nonnegative k ≤ n?

(b) Thought of as a curve in the plane, a diagonal lattice path can havemany local maxima and minima, and can have several absolute max-ima and minima, that is, several highest points and several lowestpoints. What is the y-coordinate of an absolute minimum point of aDyck Path starting at (0, 0)? Explain why a Dyck Path whose rightmostabsolute minimum point is its last point is a Catalan Path. (h)

(c) Let D be the set of all diagonal lattice paths from (0, 0) to (2n , 0). (Thusthese paths can go below the x-axis.) Suppose we partition D by lettingBi be the set of lattice paths in D that have i upsteps (perhaps mixedwith some downsteps) following the last absolute minimum. Howmany blocks does this partition have? Give a succinct description ofthe block B0. (h)

(d) How many upsteps are in a Catalan Path?

(e)∗ We are going to give a bĳection between the set of Catalan Paths andthe block Bi for each i between 1 and n. For now, suppose the valueof i, while unknown, is fixed. We take a Catalan path and break itinto three pieces. The piece F (for “front”) consists of all steps beforethe ith upstep in the Catalan path. The piece U (for “up”) consists ofthe ith upstep. The piece B (for “back”) is the portion of the path thatfollows the ith upstep. Thus we can think of the path as FUB. Showthat the function that takes FUB to BUF is a bĳection from the set ofCatalan Paths onto the block Bi of the partition. (Notice that BUF cango below the x axis.) (h)

(f) Explain how you have just given another proof of the formula for theCatalan Numbers.


aThe result we will derive is called the Chung-Feller Theorem; this approach is based of apaper of Wen-jin Woan “Uniform Partitions of Lattice Paths and Chung-Feller Generalizations,”American Mathematics Monthly 58 June/July 2001, p556.

1.3.2 The Binomial Theorem

Problem 53.◦ We know that (x + y)2 = x2 + 2x y + y2. Multiply both sidesby (x + y) to get a formula for (x + y)3 and repeat to get a formula for(x + y)4. Do you see a pattern? If so, what is it? If not, repeat the processto get a formula for (x + y)5 and look back at Figure 1.2.4 to see the pattern.Conjecture a formula for (x + y)n .

Problem 54.• When we apply the distributive law n times to (x + y)n , weget a sum of terms of the form xi yn−i for various values of the integer i. Ifit is clear to you that each term of the form xi yn−i that we get comes fromchoosing an x from i of the (x + y) factors and a y from the remaining n − iof the factors and multiplying these choices together, then answer part a ofthe problem and skip part b. In either case, be sure to answer part c.

(a) In how many ways can we choose an x from i terms and a y from n − iterms?

(b) We can take this step-by-step and consider a small case to get started.

(i) Expand the product (x1 + y1)(x2 + y2)(x3 + y3).(ii) What do you get when you substitute x for each xi and y for each

yi?(iii) Now imagine expanding

(x1 + y1)(x2 + y2) · · · (xn + yn).

Once you apply the commutative law to the individual terms youget, you will have a sum of terms of the form

xk1 xk2 · · · xki · yj1 yj2 · · · yjn−i .

What is the set {k1 , k2 , . . . , ki} ∪ { j1 , j2 , . . . , jn−i}?(iv) In how many ways can you choose the set {k1 , k2 , . . . , ki}?(v) Once you have chosen this set, how many choices do you have

for { j1 , j2 , . . . , jn−i}?(vi) If you substitute x for each xi and y for each yi , how many terms

of the form xi yn−i will you have in the expanded product

(x1 + y1)(x2 + y2) · · · (xn + yn) = (x + y)n?


(vii) How many terms of the form xn−i yi will you have?

(c) Explain how you have just proved your conjecture from Problem 53.The theorem you have proved is called the Binomial Theorem.

Problem 55. What is∑10

i=1 (10i )3

i? (h)

Problem 56. What is (n0 )− (n

1 ) + (n2 )− · · · ± (n

n ) if n is an integer bigger than0? (h)

Problem 57. Explain why

m∑i=0

(mi

) (n

k − i

)=

(m + n

k

).

Find two different explanations. (h)

Problem 58.⇒ From the symmetry of the binomial coefficients, it is not toohard to see that when n is an odd number, the number of subsets of{1, 2, . . . , n} of odd size equals the number of subsets of {1, 2, . . . , n} of evensize. Is it true that when n is even the number of subsets of {1, 2, . . . , n} ofeven size equals the number of subsets of odd size? Why or why not? (h)

Problem 59.⇒ What is∑n

i=0 i(ni )? (Hint: think about how you might use

calculus.) (h)

Notice how the proof you gave of the binomial theorem was a counting argu-ment. It is interesting that an apparently algebraic theorem that tells us how toexpand a power of a binomial is proved by an argument that amounts to count-ing the individual terms of the expansion. Part of the reason that combinatorialmathematics turns out to be so useful is that counting arguments often underlieimportant results of algebra. As the algebra becomes more sophisticated, so do thefamilies of objects we have to count, but nonetheless we can develop a great dealof algebra on the basis of counting.


1.3.3 The pigeonhole principle

Problem 60.◦ American coins are all marked with the year in which theywere made. How many coins do you need to have in your hand to guaranteethat on two (at least) of them, the date has the same last digit? (When wesay “to guarantee that on two (at least) of them,. . . ” we mean that you canfind two with the same last digit. You might be able to find three with thatlast digit, or you might be able to find one pair with the last digit 1 and onepair with the last digit 9, or any combination of equal last digits, as long asthere is at least one pair with the same last digit.)

There are many ways in which you might explain your answer to Problem 60.For example, you can partition the coins according to the last digit of their date;that is, you put all the coins with a given last digit in a block together, and putno other coins in that block; repeating until all coins are in some block. Then youhave a partition of your set of coins. If no two coins have the same last digit, theneach block has exactly one coin. Since there are only ten digits, there are at mostten blocks and so by the sum principle there are at most ten coins. In fact with tencoins it is possible to have no two with the same last digit, but with 11 coins someblock must have at least two coins in order for the sum of the sizes of at most tenblocks to be 11. This is one explanation of why we need 11 coins in Problem 60.This kind of situation arises often in combinatorial situations, and so rather thanalways using the sum principle to explain our reasoning , we enunciate anotherprinciple which we can think of as yet another variant of the sum principle. Thepigeonhole principle states that

If we partition a set with more than n elements into n parts, then at leastone part has more than one element.

The pigeonhole principle gets its name from the idea of a grid of little boxes thatmight be used, for example, to sort mail, or as mailboxes for a group of people in anoffice. The boxes in such grids are sometimes called pigeonholes in analogy withstacks of boxes used to house homing pigeons when homing pigeons were used tocarry messages. People will sometimes state the principle in a more colorful wayas “if we put more than n pigeons into n pigeonholes, then some pigeonhole hasmore than one pigeon.”

Problem 61. Show that if we have a function from a set of size n to a set ofsize less than n, then f is not one-to-one. (h)

Problem 62.• Show that if S and T are finite sets of the same size, then afunction f from S to T is one-to-one if and only if it is onto. (h)


Problem 63.· There is a generalized pigeonhole principle which says thatif we partition a set with more than kn elements into n blocks, then at leastone block has at least k + 1 elements. Prove the generalized pigeonholeprinciple. (h)

Problem 64. All the powers of five end in a five, and all the powers of twoare even. Show that for for some integer n, if you take the first n powers ofa prime other than two or five, one must have “01” as the last two digits. (h)

Problem 65.⇒ · Show that in a set of six people, there is a set of at least threepeople who all know each other, or a set of at least three people none ofwhom know each other. (We assume that if person 1 knows person 2, thenperson 2 knows person 1.) (h)

Problem 66.· Draw five circles labeled Al, Sue, Don, Pam, and Jo. Find away to draw red and green lines between people so that every pair of peopleis joined by a line and there is neither a triangle consisting entirely of redlines or a triangle consisting of green lines. What does Problem 65 tell youabout the possibility of doing this with six people’s names? What does thisproblem say about the conclusion of Problem 65 holding when there arefive people in our set rather than six?

1.3.4 Ramsey NumbersProblems 65–66 together show that six is the smallest number R with the propertythat if we have R people in a room, then there is either a set of (at least) threemutual acquaintances or a set of (at least) three mutual strangers. Another way tosay the same thing is to say that six is the smallest number so that no matter howwe connect 6 points in the plane (no three on a line) with red and green lines, wecan find either a red triangle or a green triangle. There is a name for this property.The Ramsey Number R(m , n) is the smallest number R so that if we have R peoplein a room, then there is a set of at least m mutual acquaintances or at least n mutualstrangers. There is also a geometric description of Ramsey Numbers; it uses theidea of a complete graph on R vertices. A complete graph on R vertices consists ofR points in the plane together with line segments (or curves) connecting each twoof the R vertices.1 The points are called vertices and the line segments are callededges. In Figure 1.3.2 we show three different ways to draw a complete graph onfour vertices. We use Kn to stand for a complete graph on n vertices.

1As you may have guessed, a complete graph is a special case of something called a graph. The wordgraph will be defined in Subsection 2.3.1.


Figure 1.3.2: Three ways to draw a complete graph on four vertices

Our geometric description of R(3, 3) may be translated into the language ofgraph theory (which is the subject that includes complete graphs) by saying R(3, 3)is the smallest number R so that if we color the edges of a KR with two colors,then we can find in our picture a K3 all of whose edges have the same color. Thegraph theory description of R(m , n) is that R(m , n) is the smallest number R sothat if we color the edges of a KR with red and green, then we can find in ourpicture either a Km all of whose edges are red or a Kn all of whose edges are green.Because we could have said our colors in the opposite order, we may conclude thatR(m , n) = R(n ,m). In particular R(n , n) is the smallest number R such that if wecolor the edges of a KR with two colors, then our picture contains a Kn all of whoseedges have the same color.

Problem 67.◦ Since R(3, 3) = 6, an uneducated guess might be that R(4, 4) =8. Show that this is not the case. (h)

Problem 68.· Show that among ten people, there are either four mutualacquaintances or three mutual strangers. What does this say about R(4, 3)? (h)

Problem 69.· Show that among an odd number of people there is at least oneperson who is an acquaintance of an even number of people and thereforealso a stranger to an even number of people. (h)

Problem 70.· Find a way to color the edges of a K8 with red and green sothat there is no red K4 and no green K3. (h)

Problem 71.⇒ · Find R(4, 3). (h)

As of this writing, relatively few Ramsey Numbers are known. R(3, n) is knownfor n < 10, R(4, 4) = 18, and R(5, 4) = R(4, 5) = 25.


1.4 Supplementary Chapter Problems1.⇒ Remember that we can write n as a sum of n ones. How many plus signs do weuse? In how many ways may we write n as a sum of a list of k positive numbers?Such a list is called a composition of n into k parts.

2. In Problem 1.4.1 we defined a composition of n into k parts. What is the totalnumber of compositions of n (into any number of parts).

3.· Write down a list of all 16 zero-one sequences of length four starting with 0000in such a way that each entry differs from the previous one by changing just onedigit. This is called a Gray Code. That is, a Gray Code for 0-1 sequences of length nis a list of the sequences so that each entry differs from the previous one in exactlyone place. Can you describe how to get a Gray Code for 0-1 sequences of lengthfive from the one you found for sequences of length 4? Can you describe how toprove that there is a Gray code for sequences of length n?

4.⇒ Use the idea of a Gray code from Problem 1.4.3 to prove bĳectively that thenumber of even-sized subsets of an n-element set equals the number of odd-sizedsubsets of an n-element set.

5.⇒ A list of parentheses is said to be balanced if there are the same number of leftparentheses as right, and as we count from left to right we always find at least asmany left parentheses as right parentheses. For example, (((()()))()) is balanced and((()) and (()()))(() are not. How many balanced lists of n left and n right parenthesesare there?

6.∗ Suppose we plan to put six distinct computers in a network as shown in Fig-ure 1.4.1. The lines show which computers can communicate directly with whichothers. Consider two ways of assigning computers to the nodes of the networkdifferent if there are two computers that communicate directly in one assignmentand that don’t communicate directly in the other. In how many different ways canwe assign computers to the network?

Figure 1.4.1: A computer network.

7.⇒ In a circular ice cream dish we are going to put four distinct scoops of ice creamchosen from among twelve flavors. Assuming we place four scoops of the samesize as if they were at the corners of a square, and recognizing that moving the dish

1.4. Supplementary Chapter Problems 29

doesn’t change the way in which we have put the ice cream into the dish, in howmany ways may we choose the ice cream and put it into the dish?

8.⇒ In as many ways as you can, show that (nk )(

n−km ) = (n

m )(n−mk ).

9.⇒ A tennis club has 4n members. To specify a doubles match, we choose twoteams of two people. In how many ways may we arrange the members into doublesmatches so that each player is in one doubles match? In how many ways may wedo it if we specify in addition who serves first on each team?

10. A town has n streetlights running along the north side of main street. Thepoles on which they are mounted need to be painted so that they do not rust. Inhow many ways may they be painted with red, white, blue, and green if an evennumber of them are to be painted green?

11.∗ We have n identical ping-pong balls. In how many ways may we paint themred, white, blue, and green?

12.∗ We have n identical ping-pong balls. In how many ways may we paint themred, white, blue, and green if we use green paint on an even number of them?

Chapter 2

Applications of Induction andRecursion in Combinatoricsand Graph Theory

2.1 Some Examples of Mathematical InductionIf you are unfamiliar with the Principle of Mathematical Induction, you shouldread Appendix B (a portion of which is repeated here).

2.1.1 Mathematical inductionThe principle of mathematical induction states that

In order to prove a statement about an integer n, if we can

1. Prove the statement when n = b, for some fixed integer b2. Show that the truth of the statement for n = k −1 implies the truth

of the statement for n = k whenever k > b,

then we can conclude the statement is true for all integers n ≥ b.

As an example, let us give yet another proof that a set with n elements has 2n

subsets. This proof uses essentially the same bĳections we used in proving thePascal Equation. The statement we wish to prove is the statement that “A set ofsize n has 2n subsets.”

Our statement is true when n = 0, because a set of size 0 is the empty setand the empty set has 1 = 20 subsets. (This step of our proof is called abase step.) Now suppose that k > 0 and every set with k − 1 elementshas 2k−1 subsets. Suppose S = {a1 , a2 , . . . ak} is a set with k elements.We partition the subsets of S into two blocks. Block B1 consists of thesubsets that do not contain an and block B2 consists of the subsets thatdo contain an . Each set in B1 is a subset of {a1 , a2 , . . . ak−1}, and eachsubset of {a1 , a2 , . . . ak−1} is in B1. Thus B1 is the set of all subsets of

31

32 2. Applications of Induction and Recursion in Combinatorics and Graph Theory

{a1 , a2 , . . . ak−1}. Therefore by our assumption in the first sentence ofthis paragraph, the size of B1 is 2k−1. Consider the function from B2

to B1 which takes a subset of S including ak and removes ak from it.This function is defined on B2, because every set in B2 contains ak . Thisfunction is onto, because if T is a set in B1, then T ∪ {ak} is a set in B2

which the function sends to T. This function is one-to-one because if Vand W are two different sets in B2, then removing ak from them givestwo different sets in B1. Thus we have a bĳection between B1 and B2,so B1 and B2 have the same size. Therefore by the sum principle thesize of B1 ∪ B2 is 2k−1 + 2k−1 = 2k . Therefore S has 2k subsets. Thisshows that if a set of size k − 1 has 2k−1 subsets, then a set of size k has2k subsets. Therefore by the principle of mathematical induction, a setof size n has 2n subsets for every nonnegative integer n.

The first sentence of the last paragraph is called the inductive hypothesis. In aninductive proof we always make an inductive hypothesis as part of proving thatthe truth of our statement when n = k − 1 implies the truth of our statement whenn = k. The last paragraph itself is called the inductive step of our proof. In aninductive step we derive the statement for n = k from the statement for n = k − 1,thus proving that the truth of our statement when n = k − 1 implies the truth ofour statement when n = k. The last sentence in the last paragraph is called theinductive conclusion. All inductive proofs should have a base step, an inductivehypothesis, an inductive step, and an inductive conclusion.

There are a couple details worth noticing. First, in this problem, our base stepwas the case n = 0, or in other words, we had b = 0. However, in other proofs, bcould be any integer, positive, negative, or 0. Second, our proof that the truth of ourstatement for n = k − 1 implies the truth of our statement for n = k required thatk be at least 1, so that there would be an element ak we could take away in orderto describe our bĳection. However, condition (2) of the principle of mathematicalinduction only requires that we be able to prove the implication for k > 0, so wewere allowed to assume k > 0.

2.1.1.1 Strong Mathematical Induction

One way of looking at the principle of mathematical induction is that it tells usthat if we know the “first” case of a theorem and we can derive each other case ofthe theorem from a smaller case, then the theorem is true in all cases. Howeverthe particular way in which we stated the theorem is rather restrictive in thatit requires us to derive each case from the immediately preceding case. Thisrestriction is not necessary, and removing it leads us to a more general statementof the principal of mathematical induction which people often call the strongprinciple of mathematical induction. It states:

In order to prove a statement about an integer n if we can

1. prove our statement when n = b and2. prove that the statements we get with n = b, n = b+1, . . . n = k−1

imply the statement with n = k,

then our statement is true for all integers n ≥ b.

2.1. Some Examples of Mathematical Induction 33

You will find some explicit examples of the use of the strong principle of mathe-matical induction in Appendix B and will find some uses for it in this chapter.

2.1.2 Binomial Coefficients and the Binomial Theorem

Problem 72.• When we studied the Pascal Equation and subsets in Chapter 1,it may have appeared that there is no connection between the Pascal relation(n

k ) = (n−1k−1 ) + (n−1

k ) and the formula (nk ) =

n!k!(n−k)! . Of course you probably

realize you can prove the Pascal relation by substituting the values theformula gives you into the right-hand side of the equation and simplifyingto give you the left hand side. In fact, from the Pascal Relation and the factsthat (n

0 ) = 1 and (nn ) = 1, you can actually prove the formula for (n

k ) byinduction on n. Do so. (h)

Problem 73.⇒ Use the fact that (x + y)n = (x + y)(x + y)n−1 to give aninductive proof of the binomial theorem. (h)

Problem 74. Suppose that f is a function defined on the nonnegative inte-gers such that f (0) = 3 and f (n) = 2 f (n − 1). Find a formula for f (n) andprove your formula is correct.

Problem 75. Prove the conjecture in Part 13.b for an arbitrary positive inte-ger m without appealing to the general product principle. (h)

2.1.3 Inductive definitionYou may have seen n! described by the two equations 0! = 1 and n! = n(n − 1)!for n > 0. By the principle of mathematical induction we know that this pair ofequations defines n! for all nonnegative numbers n. For this reason we call sucha definition an inductive definition. An inductive definition is sometimes calleda recursive definition. Often we can get very easy proofs of useful facts by usinginductive definitions.

Problem 76.⇒ An inductive definition of an for nonnegative n is given bya0 = 1 and an = aan−1. (Notice the similarity to the inductive definition ofn!.) We remarked above that inductive definitions often give us easy proofsof useful facts. Here we apply this inductive definition to prove two usefulfacts about exponents that you have been using almost since you learnedthe meaning of exponents.


(a) Use this definition to prove the rule of exponents am+n = am an fornonnegative m and n. (h)

(b) Use this definition to prove the rule of exponents amn = (am)n . (h)

Problem 77.+ Suppose that f is a function on the nonnegative integers suchthat f (0) = 0 and f (n) = n+ f (n−1). Prove that f (n) = n(n+1)/2. Noticethat this gives a third proof that 1 + 2 + · · · + n = n(n + 1)/2, because thissum satisfies the two conditions for f . (The sum has no terms and is thus 0when n = 0.)

Problem 78.⇒ Give an inductive definition of the summation notation∑ni=1 ai . Use it and the distributive law b(a + c) = ba + bc to prove the

distributive law

bn∑

i=1

ai =n∑

i=1

bai .

2.1.4 Proving the general product principle (Optional)We stated the sum principle as

If we have a partition of a set S, then the size of S is the sum of the sizesof the blocks of the partition.

In fact, the simplest form of the sum principle says that the size of the sum of twodisjoint (finite) sets is the sum of their sizes.

Problem 79. Prove the sum principle we stated for partitions of a set fromthe simplest form of the sum principle. (h)

We stated the simplest form of the product principle as

If we have a partition of a set S into m blocks, each of size n, then S hassize mn.

In Problem 14 we gave a more general form of the product principle which can bestated as

Let S be a set of functions f from [n] to some set X. Suppose that

• there are k1 choices for f (1), and• suppose that for each choice of f (1), f (2), . . . f (i − 1), there are ki

choices for f (i).

Then the number of functions in the set S is k1k2 · · · kn .

2.1. Some Examples of Mathematical Induction 35

Problem 80.+ Prove the general form of the product principle from the sim-plest form of the product principle. (h)

2.1.5 Double Induction and Ramsey NumbersIn Section 1.3.4 we gave two different descriptions of the Ramsey number R(m , n).However if you look carefully, you will see that we never showed that Ramseynumbers actually exist; we merely described what they were and showed thatR(3, 3) and R(3, 4) exist by computing them directly. As long as we can show thatthere is some number R such that when there are R people together, there areeither m mutual acquaintances or n mutual strangers, this shows that the RamseyNumber R(m , n) exists, because it is the smallest such R. Mathematical inductionallows us to show that one such R is (m+n−2

m−1 ). The question is, what should weinduct on, m or n? In other words, do we use the fact that with (m+n−3

m−2 ) peoplein a room there are at least m − 1 mutual acquaintances or n mutual strangers, ordo we use the fact that with at least (m+n−3

n−2 ) people in a room there are at least mmutual acquaintances or at least n − 1 mutual strangers? It turns out that we useboth. Thus we want to be able to simultaneously induct on m and n. One way todo that is to use yet another variation on the principle of mathematical induction,the Principle of Double Mathematical Induction. This principle (which can bederived from one of our earlier ones) states that

In order to prove a statement about integers m and n, if we can

1. Prove the statement when m = a and n = b, for fixed integers aand b

2. Prove the statement when m = a and n > b and when m > a andn = b (for the same fixed integers a and b),

3. Show that the truth of the statement for m = j and n = k − 1 (withj ≥ a and k > b) and the truth of the statement for m = j − 1 andn = k (with j > a and k ≥ b) imply the truth of the statement form = j and n = k,

then we can conclude the statement is true for all pairs of integers m ≥ aand n ≥ b.

There is a strong version of double induction, and it is actually easier to state. Theprinciple of strong double mathematical induction says the following.

In order to prove a statement about integers m and n, if we can

1. Prove the statement when m = a and n = b, for fixed integers aand b.

2. Show that the truth of the statemetn for values of m and n witha + b ≤ m + n < k imples the truth of the statment for m + n = k,

then we can conclude that the statement is true for all pairs of integersm ≥ a and n ≥ b.


Problem 81.⇒ · Prove that R(m , n) exists by proving that if there are (m+n−2m−1 )

people in a room, then there are either at least m mutual acquaintances orat least n mutual strangers. (h)

Problem 82.· Prove that R(m , n) ≤ R(m − 1, n) + R(m , n − 1). (h)

Problem 83.⇒ ·(a) What does the equation in Problem 82 tell us about R(4, 4)?

(b)∗ Consider 17 people arranged in a circle such that each person is ac-quainted with the first, second, fourth, and eighth person to the rightand the first, second, fourth, and eighth person to the left. can youfind a set of four mutual acquaintances? Can you find a set of fourmutual strangers? (h)

(c) What is R(4, 4)?

Problem 84. (Optional) Prove the inequality of Problem 81 by induction onm + n.

Problem 85. Use Stirling’s approximation (Problem 46) to convert the upperbound for R(n , n) that you get from Problem 81 to a multiple of a power ofan integer.

2.1.6 A bit of asymptotic combinatoricsProblem 83 gives us an upper bound on R(n , n). A very clever technique due toPaul Erdös, called the “probabilistic method,” will give a lower bound. Since bothbounds are exponential in n, they show that R(n , n) grows exponentially as n getslarge. An analysis of what happens to a function of n as n gets large is usually calledan asymptotic analysis. The probabilistic method, at least in its simpler forms,can be expressed in terms of averages, so one does not need to know the languageof probability in order to understand it. We will apply it to Ramsey numbers inthe next problem. Combined with the result of Problem 83, this problem will giveus that

√2

n< R(n , n) < 22n−2, so that we know that the Ramsey number R(n , n)

grows exponentially with n.

2.2. Recurrence Relations 37

Problem 86.⇒ Suppose we have two numbers n and m. We consider allpossible ways to color the edges of the complete graph Km with two colors,say red and blue. For each coloring, we look at each n-element subset N ofthe vertex set M of Km . Then N together with the edges of of Km connectingvertices in N forms a complete graph on n vertices. This graph, which wedenote by KN , has its edges colored by the original coloring of the edges ofKm .

(a) Why is it that if there is no subset N ⊆ M so that all the edges of KNare colored the same color, then R(n , n) > m? (h)

(b) To apply the probabilistic method, we are going to compute the av-erage, over all colorings of Km , of the number of sets N ⊆ M with|N | = n such that KN does have all its edges the same color. Explainwhy it is that if the average is less than 1, then for some coloring thereis no set N such that KN has all its edges colored the same color. Whydoes this mean that R(n , n) > m? (h)

(c) We call a KN monochromatic for a coloring c of Km if the color c(e)assigned to edge e is the same for every edge e of KN . Let us definemono(c ,N) to be 1 if N is monochromatic for c and to be 0 otherwise.Find a formula for the average number of monochromatic KNs over allcolorings of Km that involves a double sum first over all edge coloringsc of Km and then over all n-element subsets N ⊆ M of mono(c ,N). (h)

(d) Show that your formula for the average reduces to 2(mn ) · 2−(

n2)

(h)

(e) Explain why R(n , n) > m if (mn ) ≤ 2(

n2)−1. (h)

(f)∗ Explain why R(n , n) >n√

n!2(n2)−1. (h)

(g) By using Stirling’s formula, show that if n is large enough, thenR(n , n) >

√2n =

√2

n . (Here large enough means large enough forStiirling’s formula to be reasonable accurate.)

2.2 Recurrence Relations

Problem 87. How is the number of subsets of an n-element set related tothe number of subsets of an (n−1)-element set? Prove that you are correct. (h)

Problem 88. Explain why it is that the number of bĳections from an n-element set to an n-element set is equal to n times the number of bĳectionsfrom an (n − 1)-element subset to an (n − 1)-element set. What does thishave to do with Problem 27?


We can summarize these observations as follows. If sn stands for the numberof subsets of an n-element set, then

sn = 2sn−1 , (2.1)

and if bn stands for the number of bĳections from an n-element set to an n-elementset, then

bn = nbn−1. (2.2)

Equations (2.1) and (2.2) are examples of recurrence equations or recurrencerelations. A recurrence relation or simply a recurrence is an equation that ex-presses the nth term of a sequence an in terms of values of ai for i < n. ThusEquations (2.1) and (2.2) are examples of recurrences.

2.2.1 Examples of recurrence relationsOther examples of recurrences are

an = an−1 + 7, (2.3)

an = 3an−1 + 2n , (2.4)

an = an−1 + 3an−2 , and (2.5)

an = a1an−1 + a2an−2 + · · ·+ an−1a1. (2.6)

A solution to a recurrence relation is a sequence that satisfies the recurrencerelation. Thus a solution to Recurrence (2.1) is sn = 2n . Note that sn = 17 · 2n

and sn = −13 · 2n are also solutions to Recurrence (2.1). What this shows is that arecurrence can have infinitely many solutions. In a given problem, there is generallyone solution that is of interest to us. For example, if we are interested in the numberof subsets of a set, then the solution to Recurrence (2.1) that we care about is sn = 2n .Notice this is the only solution we have mentioned that satisfies s0 = 1.

Problem 89. Show that there is only one solution to Recurrence (2.1) thatsatisfies s0 = 1.

Problem 90. A first-order recurrence relation is one which expresses an interms of an−1 and other functions of n, but which does not include any ofthe terms ai for i < n − 1 in the equation.

(a) Which of the recurrences (2.1) through (2.6) are first order recurrences?

(b) Show that there is one and only one sequence an that is defined forevery nonnegative integer n, satisfies a given first order recurrence,and satisfies a0 = a for some fixed constant a. (h)

2.2. Recurrence Relations 39

Figure 2.2.1: The Towers of Hanoi Puzzle

Problem 91.⇒ The “Towers of Hanoi” puzzle has three rods rising from arectangular base with n rings of different sizes stacked in decreasing orderof size on one rod. A legal move consists of moving a ring from one rod toanother so that it does not land on top of a smaller ring. If mn is the numberof moves required to move all the rings from the initial rod to another rodthat you choose, give a recurrence for mn . (Hint: suppose you already knewthe number of moves needed to solve the puzzle with n − 1 rings.) (h)

Problem 92.⇒ We draw n mutually intersecting circles in the plane so thateach one crosses each other one exactly twice and no three intersect inthe same point. (As examples, think of Venn diagrams with two or threemutually intersecting sets.) Find a recurrence for the number rn of regionsinto which the plane is divided by n circles. (One circle divides the planeinto two regions, the inside and the outside.) Find the number of regionswith n circles. For what values of n can you draw a Venn diagram showingall the possible intersections of n sets using circles to represent each of thesets? (h)

2.2.2 Arithmetic Series (optional)

Problem 93. A child puts away two dollars from her allowance each week.If she starts with twenty dollars, give a recurrence for the amount an ofmoney she has after n weeks and find out how much money she has at theend of n weeks.

Problem 94. A sequence that satisfies a recurrence of the form an = an−1+ cis called an arithmetic progression. Find a formula in terms of the initialvalue a0 and the common difference c for the term an in an arithmeticprogression and prove you are right.

Problem 95. A person who is earning $50,000 per year gets a raise of $3000a year for n years in a row. Find a recurrence for the amount an of moneythe person earns over n + 1 years. What is the total amount of money that


the person earns over a period of n + 1 years? (In n + 1 years, there are nraises.)

Problem 96. An arithmetic series is a sequence sn equal to the sum of theterms a0 through an of of an arithmetic progression. Find a recurrence forthe sum sn of an arithmetic progression with initial value a0 and commondifference c (using the language of Problem 94). Find a formula for generalterm sn of an arithmetic series.

2.2.3 First order linear recurrencesRecurrences such as those in Equations (2.1) through (2.5) are called linear recur-rences, as are the recurrences of Problems 91 and Problem 92. A linear recurrenceis one in which an is expressed as a sum of functions of n times values of (some ofthe terms) ai for i < n plus (perhaps) another function (called the driving function)of n. A linear equation is called homogeneous if the driving function is zero (or, inother words, there is no driving function). It is called a constant coefficient linearrecurrence if the functions that are multiplied by the ai terms are all constants (butthe driving function need not be constant).

Problem 97. Classify the recurrences in Equations (2.1) through (2.5) andProblems 91 and Problem 92 according to whether or not they are constantcoefficient, and whether or not they are homogeneous.

Problem 98.• As you can see from Problem 97 some interesting sequencessatisfy first order linear recurrences, including many that have constant co-efficients, have constant driving term, or are homogeneous. Find a formulain terms of b, d, a0 and n for the general term an of a sequence that satisfiesa constant coefficient first order linear recurrence an = ban−1 + d and proveyou are correct. If your formula involves a summation, try to replace thesummation by a more compact expression. (h)

2.2.4 Geometric SeriesA sequence that satisfies a recurrence of the form an = ban−1 is called a geometricprogression. Thus the sequence satisfying Equation (2.1), the recurrence for thenumber of subsets of an n-element set, is an example of a geometric progression.From your solution to Problem 98, a geometric progression has the form an = a0bn .In your solution to Problem 98 you may have had to deal with the sum of a geometricprogression in just slightly different notation, namely

∑n−1i=0 dbi . A sum of this form

is called a (finite) geometric series.

2.3. Graphs and Trees 41

Problem 99. Do this problem only if your final answer (so far) to Problem 98contained the sum

∑n−1i=0 dbi .

(a) Expand (1− x)(1+ x). Expand (1− x)(1+ x + x2). Expand (1− x)(1+x + x2 + x3).

(b) What do you expect (1− b)∑n−1

i=0 dbi to be? What formula for∑n−1

i=0 dbi

does this give you? Prove that you are correct.

In Problem 98 and perhaps 99 you proved an important theorem.

Theorem 2.2.2. If b " 1 and an = ban−1 + d, then an = a0bn + d1 − bn

1 − b. If b = 1,

then, an = a0 + nd

Corollary 2.2.3. If b " 1, thenn−1∑i=0

bi =1 − bn

1 − b. If b = 1,

n−1∑i=0

bi = n.

2.3 Graphs and Trees2.3.1 Undirected graphsIn Section 1.3.4 we introduced the idea of a directed graph. Graphs consist ofvertices and edges. We describe vertices and edges in much the same way as wedescribe points and lines in geometry: we don’t really say what vertices and edgesare, but we say what they do. We just don’t have a complicated axiom system theway we do in geometry. A graph consists of a set V called a vertex set and a setE called an edge set. Each member of V is called a vertex and each member ofE is called an edge. Associated with each edge are two (not necessarily different)vertices called its endpoints. We draw pictures of graphs by drawing points torepresent the vertices and line segments (curved if we choose) whose endpointsare at vertices to represent the edges. In Figure 2.3.1 we show three pictures ofgraphs.


z

w

xy

v

ab

c

d

e

f

1

2

3

4

5

6

7

8

Figure 2.3.1: Three different graphs

Each gray circle in the figure represents a vertex; each line segment representsan edge. You will note that we labelled the vertices; these labels are names we choseto give the vertices. We can choose names or not as we please. The third graphalso shows that it is possible to have an edge that connects a vertex (like the onelabelled y) to itself or it is possible to have two or more edges (like those betweenvertices v and y) between two vertices. The degree of a vertex is the number oftimes it appears as the endpoint of edges; thus the degree of y in the third graphin the figure is four.

Problem 100.◦ In the graph on the left in Figure 2.3.1, what is the degree ofeach vertex?

Problem 101.◦ For each graph in Figure 2.3.1 is the number of vertices ofodd degree even or odd?

Problem 102.⇒ · The sum of the degrees of the vertices of a (finite) graph isrelated in a natural way to the number of edges.

(a) What is the relationship? (h)

(b) Find a proof that what you say is correct that uses induction on thenumber of edges. Hint: To make your inductive step, think aboutwhat happens to a graph if you delete an edge. (h)


(c) Find a proof that what you say is correct that uses induction on thenumber of vertices.

(d) Find a proof that what you say is correct that does not use induction. (h)

Problem 103.· What can you say about the number of vertices of odd degreein a graph? (h)

2.3.2 Walks and paths in graphsA walk in a graph is an alternating sequence v0e1v1 . . . ei vi of vertices and edgessuch that edge ei connects vertices vi−1 and vi . A graph is called connected if, forany pair of vertices, there is a walk starting at one and ending at the other.

Problem 104. Which of the graphs in Figure 2.3.1 is connected?

Problem 105.◦ A path in a graph is a walk with no repeated vertices. Findthe longest path you can in the third graph of Figure 2.3.1.

Problem 106.◦ A cycle in a graph is a walk whose first and last vertex areequal but which has no other repeated vertices. Which graphs in Figure 2.3.1have cycles? What is the largest number of edges in a cycle in the secondgraph in Figure 2.3.1? What is the smallest number of edges in a cycle inthe third graph in Figure 2.3.1?

Problem 107.◦ A connected graph with no cycles is called a tree. Whichgraphs, if any, in Figure 2.3.1 are trees?

2.3.3 Counting vertices, edges, and paths in trees

Problem 108.⇒ · Draw some trees and on the basis of your examples, make aconjecture about the relationship between the number of vertices and edgesin a tree. Prove your conjecture. (Hint: what happens if you choose an edgeand delete it, but not its endpoints?) (h)


Problem 109.· What is the minimum number of vertices of degree one in afinite tree? What is it if the number of vertices is bigger than one? Provethat you are correct. (h)

Problem 110.⇒ · In a tree, given two vertices, how many paths can you findbetween them? Prove that you are correct.

Problem 111.⇒ ∗ How many trees are there on the vertex set {1, 2}? On thevertex set {1, 2, 3}? When we label the vertices of our tree, we consider thetree which has edges between vertices 1 and 2 and between vertices 2 and 3different from the tree that has edges between vertices 1 and 3 and between2 and 3. See Figure 2.3.2.

1 23

2 31

2 13

Figure 2.3.2: The three labelled trees on three vertices

How many (labelled) trees are there on four vertices? You don’t have a lotof data to guess from, but try to guess a formula for the number of labelledtrees with vertex set {1, 2, · · · , n}. (h)

We are now going to introduce a method to prove the formula you guessed.Given a tree with two or more vertices, labelled with positive integers, we define asequence b1 , b2 , . . . of integers inductively as follows: If the tree has two vertices,the sequence consists of one entry, namely the label of the vertex with the largerlabel. Otherwise, let a1 be the lowest numbered vertex of degree 1 in the tree. Letb1 be the label of the unique vertex in the tree adjacent to a1 and write down b1.For example, in the first graph in Figure 2.3.1, a1 is 1 and b1 is 2. Given a1 throughai−1, let ai be the lowest numbered vertex of degree 1 in the tree you get by deletinga1 through ai−1and let bi be the unique vertex in this new tree adjacent to ai . Forexample, in the first graph in Figure 2.3.1, a2 = 2 and b2 = 3. Then a3 = 5 andb3 = 4. We use b to stand for the sequence of bis we get in this way. In the tree (thefirst graph) in Figure 2.3.1, the sequence b is 2344378. (If you are unfamiliar withinductive (recursive) definition, you might want to write down some other labelledtrees on eight vertices and construct the sequence of bis.)

Problem 112.(a) How long will the sequence of bis be if it is computed from a tree with

n vertices (labelled with 1 through n)?

(b) What can you say about the last member of the sequence of bis? (h)


(c) Can you tell from the sequence of bis what a1is? (h)

(d) Find a bĳection between labelled trees and something you can “count”that will tell you how many labelled trees there are on n labelledvertices. (h)

The sequence b1 , b2 , . . . , bn−2 in Problem 111 is called a Prüfer coding or Prüfercode for the tree. There is a good bit of interesting information encoded into thePrüfer code for a tree.

Problem 113. What can you say about the vertices of degree one from thePrüfer code for a tree labeled with the integers from 1 to b? (h)

Problem 114. What can you say about the Prüfer code for a tree with exactlytwo vertices of degree 1? (and perhaps some vertices with other degrees aswell)? Does this characterize such trees?

Problem 115.⇒ What can you determine about the degree of the vertex la-belled i from the Prüfer code of the tree? (h)

Problem 116.⇒ What is the number of (labelled) trees on n vertices with threevertices of degree 1? (Assume they are labelled with the integers 1 throughn.) This problem will appear again in the next chapter after some materialthat will make it easier. (h)

2.3.4 Spanning treesMany of the applications of trees arise from trying to find an efficient way to connectall the vertices of a graph. For example, in a telephone network, at any given timewe have a certain number of wires (or microwave channels, or cellular channels)available for use. These wires or channels go from a specific place to a specificplace. Thus the wires or channels may be thought of as edges of a graph and theplaces the wires connect may be thought of as vertices of that graph. A tree whoseedges are some of the edges of a graph G and whose vertices are all of the verticesof the graph G is called a spanning tree of G. A spanning tree for a telephonenetwork will give us a way to route calls between any two vertices in the network.In Figure 2.3.3 we show a graph and all its spanning trees.


Figure 2.3.3: A graph and all its spanning trees.

Problem 117. Show that every connected graph has a spanning tree. It ispossible to find a proof that starts with the graph and works “down” towardsthe spanning tree and to find a proof that starts with just the vertices andworks “up” towards the spanning tree. Can you find both kinds of proof?

2.3.5 Minimum cost spanning treesOur motivation for talking about spanning trees was the idea of finding a min-imum number of edges we need to connect all the edges of a communicationnetwork together. In many cases edges of a communication network come withcosts associated with them. For example, one cell-phone operator charges anotherone when a customer of the first uses an antenna of the other. Suppose a companyhas offices in a number of cities and wants to put together a communication net-work connecting its various locations with high-speed computer communications,but to do so at minimum cost. Then it wants to take a graph whose vertices are thecities in which it has offices and whose edges represent possible communicationslines between the cities. Of course there will not necessarily be lines between eachpair of cities, and the company will not want to pay for a line connecting city i andcity j if it can already connect them indirectly by using other lines it has chosen.Thus it will want to choose a spanning tree of minimum cost among all spanningtrees of the communications graph. For reasons of this application, if we have agraph with numbers assigned to its edges, the sum of the numbers on the edges ofa spanning tree of G will be called the cost of the spanning tree.

Problem 118.⇒ Describe a method (or better, two methods different in at leastone aspect) for finding a spanning tree of minimum cost in a graph whoseedges are labelled with costs, the cost on an edge being the cost for includingthat edge in a spanning tree. Prove that your method(s) work. (h)

The method you used in Problem 118 is called a greedy method, because eachtime you made a choice of an edge, you chose the least costly edge available toyou.


2.3.6 The deletion/contraction recurrence for spanning treesThere are two operations on graphs that we can apply to get a recurrence (thougha more general kind than those we have studied for sequences) which will let uscompute the number of spanning trees of a graph. The operations each apply to anedge e of a graph G. The first is called deletion; we delete the edge e from the graphby removing it from the edge set. Figure 2.3.4 shows how we can delete edges froma graph to get a spanning tree.

Figure 2.3.4: Deleting two appropriate edges from this graph gives a spanning tree.

The second operation is called contraction.

ee

1 23e

E

E

E

4

56

7

1 23

4

56

7

1 23

4

56

7

13

46

7

23

4

5

7

1

4

56

7

Figure 2.3.5: The results of contracting three different edges in a graph.

Contractions of three different edges in the same graph are shown in Figure 2.3.5.Intuitively, we contract an edge by shrinking it in length until its endpoints coincide;we let the rest of the graph “go along for the ride.” To be more precise, we contractthe edge e with endpoints v and w as follows:

1. remove all edges having either v or w or both as an endpoint from the edgeset,

2. remove v and w from the vertex set,

3. add a new vertex E to the vertex set,

4. add an edge from E to each remaining vertex that used to be an endpoint ofan edge whose other endpoint was v or w, and add an edge from E to E forany edge other than e whose endpoints were in the set {v , w}.

We use G − e (read as G minus e) to stand for the result of deleting e from G,and we use G/e (read as G contract e) to stand for the result of contracting e fromG.


Problem 119.⇒ ·(a) How do the number of spanning trees of G not containing the edge

e and the number of spanning trees of G containing e relate to thenumber of spanning trees of G − e and G/e? (h)

(b) Use (G) to stand for the number of spanning trees of G (so that, forexample, (G/e) stands for the number of spanning trees of G/e).Find an expression for (G) in terms of (G/e) and (G − e). Thisexpression is called the deletion-contraction recurrence.

(c) Use the recurrence of the previous part to compute the number ofspanning trees of the graph in Figure 2.3.6.

1 2

3

4

5

Figure 2.3.6: A graph.

2.3.7 Shortest paths in graphsSuppose that a company has a main office in one city and regional offices in othercities. Most of the communication in the company is between the main office andthe regional offices, so the company wants to find a spanning tree that minimizesnot the total cost of all the edges, but rather the cost of communication between themain office and each of the regional offices. It is not clear that such a spanning treeeven exists. This problem is a special case of the following. We have a connectedgraph with nonnegative numbers assigned to its edges. (In this situation thesenumbers are often called weights.) The (weighted) length of a path in the graphis the sum of the weights of its edges. The distance between two vertices is theleast (weighted) length of any path between the two vertices. Given a vertex v, wewould like to know the distance between v and each other vertex, and we wouldlike to know if there is a spanning tree in G such that the length of the path in thespanning tree from v to each vertex x is the distance from v to x in G.

Problem 120. Show that the following algorithm (known as Dĳkstra’s algo-rithm) applied to a weighted graph whose vertices are labelled 1 to n gives,for each i, the distance from vertex 1 to i as d(i).

1. Let d(1) = 0. Let d(i) = ∞ for all other i. Let v(1)=1. Let v( j) = 0 forall other j. For each i and j, let w(i , j) be the minimum weight of an

2.4. Supplementary Problems 49

edge between i and j, or ∞ if there are no such edges. Let k = 1. Lett = 1.

2. For each i, if d(i) > d(k) + w(k , i) let d(i) = d(k) + w(k , i).

3. Among those i with v(i) = 0, choose one with d(i) a minimum, andlet k = i. Increase t by 1. Let v(i) = 1.

4. Repeat the previous two steps until t = n

Problem 121. Is there a spanning tree such that the distance from vertex 1to vertex i given by the algorithm in Problem 120 is the distance for vertex1 to vertex i in the tree (using the same weights on the edges, of course)?

2.4 Supplementary Problems1. Use the inductive definition of an to prove that (ab)n = an bn for all nonnegativeintegers n.

2. Give an inductive definition ofn⋃

i=1

Si and use it and the two set distributive law

to prove the distributive law A ∩n⋃

i=1

Si =n⋃

i=1

A ∩ Si .

3.⇒ A hydrocarbon molecule is a molecule whose only atoms are either carbon atomsor hydrogen atoms. In a simple molecular model of a hydrocarbon, a carbon atomwill bond to exactly four other atoms and hydrogen atom will bond to exactly oneother atom. Such a model is shown in Figure 2.4.1. We represent a hydrocarboncompound with a graph whose vertices are labelled with C’s and H’s so that each Cvertex has degree four and each H vertex has degree one. A hydrocarbon is calledan “alkane” Common examples are methane (natural gas), butane (one versionof which is shown in Figure 2.4.1)propane, hexane (ordinary gasoline), octane (tomake gasoline burn more slowly), etc.

Figure 2.4.1: A model of a butane molecule

(a) How many vertices are labelled H in the graph of an alkane with exactly nvertices labelled C?


(b) An alkane is called butane if it has four carbon atoms. Why do we say oneversion of butane is shown in Figure 2.4.1?

4.(a) Give a recurrence for the number of ways to divide 2n people into sets of two

for tennis games. (Don’t worry about who serves first.)(b) Give a recurrence for the number of ways to divide 2n people into sets of two

for tennis games and to determine who serves first.

5.⇒ Give a recurrence for the number of ways to divide 4n people into sets of fourfor games of bridge. (Don’t worry about how they sit around the bridge table orwho is the first dealer.)

6. Use induction to prove your result in Supplementary Problem 1.4.2 at the endof Chapter 1.

7. Give an inductive definition of the product notationn∏

i=1

ai .

8. Using the fact that (ab)k = ak bk , use your inductive definition of product nota-

tion in Problem 2.4.7 to prove that(

n∏i=1

ai

) k

=n∏

i=1

aki .

9.⇒ How many labelled trees on n vertices have exactly four vertices of degree 1?

10.⇒ ∗ The degree sequence of a tree is a list of the degrees of the vertices in nonin-creasing order. For example the degree sequence of the first graph in Figure 2.3.3is (4, 3, 2, 2, 1). For a graph with vertices labeled 1 through n, the ordered degreesequence of the graph is the sequence (d1 , d2 , . . . , dn) in which di is the degree ofvertex i. For example, the ordered degree sqeuence of the first graph in Figure 2.3.1is (1, 2, 3, 3, 1, 1, 2, 1).(a) How many labelled trees are there on n vertices with ordered degree sequence

d1 , d2 , . . . dn? (This problem appears again in the next chapter since some ideasin that chapter make it more straightforward.)

(b) How many labeled trees are there on n vertices with the degree sequence inwhich the degree d appears id times?

Chapter 3

Distribution Problems

3.1 The idea of a distributionMany of the problems we solved in Chapter 1 may be thought of as problems ofdistributing objects (such as pieces of fruit or ping-pong balls) to recipients (such aschildren). Some of the ways of viewing counting problems as distribution problemsare somewhat indirect. For example, in Problem 37 you probably noticed that thenumber of ways to pass out k ping-pong balls to n children so that no child getsmore than one is the number of ways that we may choose a k-element subset of ann-element set. We think of the children as recipients and objects we are distributingas the identical ping-pong balls, distributed so that each recipient gets at most oneball. Those children who receive an object are in our set. It is helpful to havemore than one way to think of solutions to problems. In the case of distributionproblems, another popular model for distributions is to think of putting balls inboxes rather than distributing objects to recipients. Passing out identical objectsis modeled by putting identical balls into boxes. Passing out distinct objects ismodeled by putting distinct balls into boxes.

3.1.1 The twentyfold wayWhen we are passing out objects to recipients, we may think of the objects asbeing either identical or distinct. We may also think of the recipients as beingeither identical (as in the case of putting fruit into plastic bags in the grocery store)or distinct (as in the case of passing fruit out to children). We may restrict thedistributions to those that give at least one object to each recipient, or those thatgive exactly one object to each recipient, or those that give at most one object toeach recipient, or we may have no such restrictions. If the objects are distinct, itmay be that the order in which the objects are received is relevant (think aboutputting books onto the shelves in a bookcase) or that the order in which the objectsare received is irrelevant (think about dropping a handful of candy into a child’strick or treat bag). If we ignore the possibility that the order in which objects arereceived matters, we have created 2 · 2 · 4 = 16 distribution problems. In the caseswhere a recipient can receive more than one distinct object, we also have four more

51

52 3. Distribution Problems

problems when the order objects are received matters. Thus we have 20 possibledistribution problems.

The Twentyfold Way: A Table of Distribution Problemsk objects and conditionson how they are received

n recipients and mathematicalmodel for distribution

Distinct Identical1. Distinctno conditions

nk

functions?

set partitions (≤ n parts)

2. DistinctEach gets at most one

nk

k-elementpermutations

1 if k ≤ n;0 otherwise

3. DistinctEach gets at least one

?

onto functions?

set partitions (n parts)4. DistinctEach gets exactly one

k! = n!permutations

1 if k = n;0 otherwise

5. Distinct,order matters

??

??

6. Distinct,order mattersEach gets at least one

??

??

7. Identicalno conditions

??

??

8. IdenticalEach gets at most one

(nk )

subsets1 if k ≤ n;

0 otherwise9. IdenticalEach gets at least one

??

??

10. IdenticalEach gets exactly one



Table 3.1.1: An incomplete table of the number of ways to distribute k objects to nrecipients, with restrictions on how the objects are received

We describe these problems in Table 3.1.1. Since there are twenty possible dis-tribution problems, we call the table the “Twentyfold Way,” adapting terminologysuggested by Joel Spencer for a more restricted class of distribution problems. Inthe first column of the table we state whether the objects are distinct (like people)or identical (like ping-pong balls) and then give any conditions on how the ob-jects may be received. The conditions we consider are whether each recipient getsat most one object, whether each recipient gets at least one object, whether eachrecipient gets exactly one object, and whether the order in which the objects arereceived matters. In the second column we give the solution to the problem andthe name of the mathematical model for this kind of distribution problem when therecipients are distinct, and in the third column we give the same information whenthe recipients are identical. We use question marks as the answers to problems we

3.1. The idea of a distribution 53

have not yet solved and models we have not yet studied. We give explicit answersto problems we solved in Chapter 1 and problems whose answers are immediate.The goal of this chapter is to develop methods that will allow us to fill in the tablewith formulas or at least quantities we know how to compute, and we will givea completed table at the end of the chapter. We will now justify the answers thatare not question marks and replace some question marks with answers as we coverrelevant material.

If we pass out k distinct objects (say pieces of fruit) to n distinct recipients (saychildren), we are saying for each object which recipient it goes to. Thus we aredefining a function from the set of objects to the recipients. We saw the followingtheorem in Problem 13.b.Theorem 3.1.2. There are nk functions from a k-element set to an n-element set.

We proved it in Problem 13.b and in another way in Problem 75. If we pass outk distinct objects (say pieces of fruit) to n indistinguishable recipients (say identicalpaper bags) then we are dividing the objects up into disjoint sets; that is we areforming a partition of the objects into some number, certainly no more than thenumber k of objects, of parts. Later in this chapter (and again in the next chapter)we shall discuss how to compute the number of partitions of a k-element set into nparts. This explains the entries in row one of our table.

If we pass out k distinct objects to n recipients so that each gets at most one,we still determine a function, but the function must be one-to-one. The number ofone-to-one functions from a k-element set to an n element set is the same as thenumber of one-to-one functions from the set [k] = {1, 2, . . . , k} to an n-element set.In Problem 20 we proved the following theorem.Theorem 3.1.3. If 0 ≤ k ≤ n, then the number of k-element permutations of an n-elementset is

nk = n(n − 1) · · · (n − k + 1) = n!/(n − k)!.

If k > n there are no one-to-one functions from a k element set to an n element, sowe define nk to be zero in this case. Notice that this is what the indicated productin the middle term of our formula gives us. If we are supposed to distribute kdistinct objects to n identical recipients so that each gets at most one, we cannot doso if k > n, so there are 0 ways to do so. On the other hand, if k ≤ n, then it doesn’tmatter which recipient gets which object, so there is only one way to do so. Thisexplains the entries in row two of our table.

If we distribute k distinct objects to n distinct recipients so that each recipientgets at least one, then we are counting functions again, but this time functions froma k-element set onto an n-element set. At present we do not know how to computethe number of such functions, but we will discuss how to do so later in this chapterand in the next chapter. If we distribute k identical objects to n recipients, we areagain simply partitioning the objects, but the condition that each recipient gets atleast one means that we are partitioning the objects into exactly n blocks. Again,we will discuss how compute the number of ways of partitioning a set of k objectsinto n blocks later in this chapter. This explains the entries in row three of ourtable.

If we pass out k distinct objects to n recipients so that each gets exactly one, thenk = n and the function that our distribution gives us is a bĳection. The number of


bĳections from an n-element set to an n-element set is n! by Theorem 3.1.3. If wepass out k distinct objects of n identical recipients so that each gets exactly 1, thenin this case it doesn’t matter which recipient gets which object, so the number ofways to do so is 1 if k = n. If k " n, then the number of such distributions is zero.This explains the entries in row four of our table.

We now jump to row eight of our table. We saw in Problem 37 that the numberof ways to pass out k identical ping-pong balls to n children is simply the numberof k-element subsets of an n-element set. In Problem 39 we proved the followingtheorem.Theorem 3.1.4. If 0 ≤ k ≤ n, the number of k-element subsets of an n-element set isgiven by (

nk

)=

nk

k!=

n!k!(n − k)!

.

We define (nk ) to be 0 if k > n, because then there are no k-element subsets of

an n-element set. Notice that this is what the middle term of the formula in thetheorem gives us. This explains the entries of row 8 of our table. For now we jumpover row 9.

In row 10 of our table, if we are passing out k identical objects to n recipientsso that each gets exactly one, it doesn’t matter whether the recipients are identicalor not; there is only one way to pass out the objects if k = n and otherwise it isimpossible to make the distribution, so there are no ways of distributing the objects.This explains the entries of row 10 of our table. Several other rows of our table canbe computed using the methods of Chapter 1.

3.1.2 Ordered functions

Problem 122. Suppose we wish to place k distinct books onto the shelvesof a bookcase with n shelves. For simplicity, assume for now that all of thebooks would fit on any of the shelves. Also, let’s imagine pushing the bookson a shelf as far to the left as we can, so that we are only thinking about howthe books sit relative to each other, not about the exact places where we putthe books. Since the books are distinct, we can think of a the first book, thesecond book and so on.

(a) How many places are there where we can place the first book?

(b) When we place the second book, if we decide to place it on the shelfthat already has a book, does it matter if we place it to the left or rightof the book that is already there?

(c) How many places are there where we can place the second book? (h)

(d) Once we have i − 1 books placed, if we want to place book i on a shelfthat already has some books, is sliding it in to the left of all the booksalready there different from placing it to the right of all the booksalready or between two books already there?

(e) In how many ways may we place the ith book into the bookcase? (h)


(f) In how many ways may we place all the books?

Problem 123. Suppose we wish to place the books in Problem 122 (satis-fying the assumptions we made there) so that each shelf gets at least onebook. Now in how many ways may we place the books? (Hint: how canyou make sure that each shelf gets at least one book before you start theprocess described in Problem 122?) (h)

The assignment of which books go to which shelves of a bookcase is simply afunction from the books to the shelves. But a function doesn’t determine whichbook sits to the left of which others on the shelf, and this information is part ofhow the books are arranged on the shelves. In other words, the order in whichthe shelves receive their books matters. Our function must thus assign an orderedlist of books to each shelf. We will call such a function an ordered function. Moreprecisely, an ordered function from a set S to a set T is a function that assigns an(ordered) list of elements of S to some, but not necessarily all, elements of T in sucha way that each element of S appears on one and only one of the lists.1 (Noticethat although it is not the usual definition of a function from S to T, a functioncan be described as an assignment of subsets of S to some, but not necessarily all,elements of T so that each element of S is in one and only one of these subsets.)Thus the number of ways to place the books into the bookcase is the entry in themiddle column of row 5 of our table. If in addition we require each shelf to get atleast one book, we are discussing the entry in the middle column of row 6 of ourtable. An ordered onto function is one which assigns a list to each element of T. InProblem 122 you showed that the number of ordered functions from a k-element

set to an n-element set isk∏

i=1

(n + i − 1). This product occurs frequently enough

that it has a name; it is called the kth rising factorial power of n and is denotedby nk . It is read as “n to the k rising.” (This notation is due to Don Knuth, whoalso suggested the notation for falling factorial powers.) We can summarize with atheorem that adds two more formulas for the number of ordered functions.

Theorem 3.1.5. The number of ordered functions from a k-element set to an n-element setis

nk =k∏

i=1

(n + i − 1) =(n + i − 1)!

(n − 1)!= (n + k − 1)k .

Ordered functions explain the entries in the middle column of rows 5 and 6 ofour table of distribution problems.

1The phrase ordered function is not a standard one, because there is as yet no standard name for theresult of an ordered distribution problem.


3.1.3 MultisetsIn the middle column of row 7 of our table, we are asking for the number of waysto distribute k identical objects (say ping-pong balls) to n distinct recipients (saychildren).

Problem 124.• In how many ways may we distribute k identical books onthe shelves of a bookcase with n shelves, assuming that any shelf can holdall the books? (h)

Problem 125.• A multiset chosen from a set S may be thought of as a subsetwith repeated elements allowed. For example the multiset of letters of theword Mississippi is {i , i , i , i ,m , p , p , s , s , s , s}. To determine a multiset wemust say how many times (including, perhaps, zero) each member of Sappears in the multiset. The number of times an element appears is calledits multiplicity. The size of a multiset chosen from S is the total numberof times any member of S appears. For example, the size of the multiset ofletters of Mississippi is 11. What is the number of multisets of size k thatcan be chosen from an n-element set? (h)

Problem 126.⇒ Your answer in the previous problem should be expressibleas a binomial coefficient. Since a binomial coefficient counts subsets, find abĳection between subsets of something and multisets chosen from a set S. (h)

Problem 127. How many solutions are there in nonnegative integers to theequation x1 + x2 + · · ·+ xm = r, where m and r are constants? (h)

Problem 128. In how many ways can we distribute k identical objects to ndistinct recipients so that each recipient gets at least m? (h)

Multisets explain the entry in the middle column of row 7 of our table ofdistribution problems.

3.1.4 Compositions of integers

Problem 129.· In how many ways may we put k identical books onto nshelves if each shelf must get at least one book? (h)


Problem 130.· A composition of the integer k into n parts is a list of npositive integers that add to k. How many compositions are there of aninteger k into n parts? (h)

Problem 131.⇒ Your answer in Problem 130 can be expressed as a binomialcoefficient. This means it should be possible to interpret a composition asa subset of some set. Find a bĳection between compositions of k into nparts and certain subsets of some set. Explain explicitly how to get thecomposition from the subset and the subset from the composition. (h)

Problem 132.· Explain the connection between compositions of k into n partsand the problem of distributing k identical objects to n recipients so thateach recipient gets at least one.

The sequence of problems you just completed should explain the entry in themiddle column of row 9 of our table of distribution problems.

3.1.5 Broken permutations and Lah numbers

Problem 133.⇒ · In how many ways may we stack k distinct books into nidentical boxes so that there is a stack in every box? The hints may suggestthat you can do this problem in more than one way! (h)

We can think of stacking books into identical boxes as partitioning the booksand then ordering the blocks of the partition. This turns out not to be a usefulcomputational way of visualizing the problem because the number of ways toorder the books in the various stacks depends on the sizes of the stacks and notjust the number of stacks. However this way of thinking actually led to the firsthint in Problem 133. Instead of dividing a set up into nonoverlapping parts, wemay think of dividing a permutation (thought of as a list) of our k objects up inton ordered blocks. We will say that a set of ordered lists of elements of a set S is abroken permutation of S if each element of S is in one and only one of these lists.2The number of broken permutations of a k-element set with n blocks is denotedby L(k , n). The number L(k , n) is called a Lah Number and, from our solution toProblem 133, is equal to k!(k−1

n−1 )/n!.The Lah numbers are the solution to the question “In how many ways may

we distribute k distinct objects to n identical recipients if order matters and eachrecipient must get at least one?" Thus they give the entry in row 6 and column 3of our table. The entry in row 5 and column 3 of our table will be the number of

2The phrase broken permutation is not standard, because there is no standard name for the solutionto this kind of distribution problem.


broken permutations with less than or equal to n parts. Thus it is a sum of Lahnumbers.

We have seen that ordered functions and broken permutations explain theentries in rows 5 and 6 of our table.

In the next two sections we will give ways of computing the remaining entries.

3.2 Partitions and Stirling NumbersWe have seen how the number of partitions of a set of k objects into n blockscorresponds to the distribution of k distinct objects to n identical recipients. Whilethere is a formula that we shall eventually learn for this number, it requires moremachinery than we now have available. However there is a good method forcomputing this number that is similar to Pascal’s equation. Now that we havestudied recurrences in one variable, we will point out that Pascal’s equation is infact a recurrence in two variables; that is it lets us compute (n

k ) in terms of values of(m

i ) in which either m < n or i < k or both. It was the fact that we had such arecurrence and knew (n

0 ) and (nn ) that let us create Pascal’s triangle.

3.2.1 Stirling Numbers of the second kindWe use the notation S(k , n) to stand for the number of partitions of a k elementset with n blocks. For historical reasons, S(k , n) is called a Stirling number of thesecond kind.

Problem 134. In a partition of the set [k], the number k is either in a blockby itself, or it is not. How does the number of partitions of [k] with n partsin which k is in a block with other elements of [k] compare to the number ofpartitions of [k −1] into n blocks? Find a two variable recurrence for S(n , k),valid for n and k larger than one. (h)

Problem 135. What is S(k , 1)? What is S(k , k)? Create a table of values ofS(k , n) for k between 1 and 5 and n between 1 and k. This table is sometimescalled Stirling’s Triangle (of the second kind) How would you defineS(k , n) for the nonnegative values of k and n that are not both positive?Now for what values of k and n is your two variable recurrence valid?

Problem 136. Extend Stirling’s triangle enough to allow you to answer thefollowing question and answer it. (Don’t fill in the rows all the way; thework becomes quite tedious if you do. Only fill in what you need to answerthis question.) A caterer is preparing three bag lunches for hikers. Thecaterer has nine different sandwiches. In how many ways can these nine

3.2. Partitions and Stirling Numbers 59

sandwiches be distributed into three identical lunch bags so that each baggets at least one?

Problem 137. The question in Problem 136 naturally suggests a more re-alistic question; in how many ways may the caterer distribute the ninesandwich’s into three identical bags so that each bag gets exactly three?Answer this question. (h)

Problem 138.· What is S(k , k − 1)? (h)

Problem 139.• In how many ways can we partition k items into n blocks sothat we have ki blocks of size i for each i? (Notice that

∑ki=1 ki = n and∑k

i=1 iki = k.) The sequence k1 , k2 , . . . , kn is called the type vector of thepartition. (h)

Problem 140.+ Describe how to compute S(k , n) in terms of quantities givenby the formula you found in Problem 139.

Problem 141.⇒ Find a recurrence for the Lah numbers L(k , n) similar to theone in Problem 134. (h)

Problem 142.· (Relevant in Appendix C.) The total number of partitions ofa k-element set is denoted by B(k) and is called the k-th Bell number. ThusB(1) = 1 and B(2) = 2.

(a) Show, by explicitly exhibiting the partitions, that B(3) = 5.

(b) Find a recurrence that expresses B(k) in terms of B(n) for n < k andprove your formula correct in as many ways as you can. (h)

(c) Find B(k) for k = 4, 5, 6.


3.2.2 Stirling Numbers and onto functions

Problem 143.◦ Given a function f from a k-element set K to an n-elementset, we can define a partition of K by putting x and y in the same block of thepartition if and only if f (x) = f (y). How many blocks does the partitionhave if f is onto? How is the number of functions from a k-element set ontoan n-element set related to a Stirling number? Be as precise in your answeras you can. (h)

Problem 144.⇒ How many labeled trees on n vertices have exactly 3 verticesof degree one? Note that this problem has appeared before in Chapter 2. (h)

Problem 145.• Each function from a k-element set K to an n-element set Nis a function from K onto some subset of N . If J is a subset of N of size j, youknow how to compute the number of functions that map onto J in terms ofStirling numbers. Suppose you add the number of functions mapping ontoJ over all possible subsets J of N . What simple value should this sum equal?Write the equation this gives you. (h)

Problem 146.◦ In how many ways can the sandwiches of Problem 136 beplaced into three distinct bags so that each bag gets at least one?

Problem 147.◦ In how many ways can the sandwiches of Problem 137 beplaced into distinct bags so that each bag gets exactly three?

Problem 148.• In how many ways may we label the elements of a k-elementset with n distinct labels (numbered 1 through n) so that label i is used jitimes? (If we think of the labels as y1 , y2 , . . . , yn , then we can rephrase thisquestion as follows. How many functions are there from a k-element set Kto a set N = {y1 , y2 , . . . yn} so that yi is the image of ji elements of K?) Thisnumber is called a multinomial coefficient and denoted by

(k

j1 , j2 , . . . , jn

).

(h)

3.2. Partitions and Stirling Numbers 61

Problem 149. Explain how to compute the number of functions from a k-element set K to an n-element set N by using multinomial coefficients. (h)

Problem 150. Explain how to compute the number of functions from a k-element set K onto an n-element set N by using multinomial coefficients. (h)

Problem 151.• What do multinomial coefficients have to do with expandingthe kth power of a multinomial x1 + x2 + · · · + xn? This result is called themultinomial theorem. (h)

3.2.3 Stirling Numbers and bases for polynomials

Problem 152.·(a) Find a way to express nk in terms of k j for appropriate values j. You

may use Stirling numbers if they help you. (h)

(b) Notice that x j makes sense for a numerical variable x (that could rangeover the rational numbers, the real numbers, or even the complexnumbers instead of only the nonnegative integers, as we are implicitlyassuming n does), just as x j does. Find a way to express the power xk

in terms of the polynomials x j for appropriate values of j and explainwhy your formula is correct. (h)

You showed in Problem 152 how to get each power of x in terms of the fallingfactorial powers x j . Therefore every polynomial in x is expressible in terms of asum of numerical multiples of falling factorial powers. Using the language of linearalgebra, we say that the ordinary powers of x and the falling factorial powers of xeach form a basis for the “space” of polynomials, and that the numbers S(k , n) are“change of basis coefficients.” If you are not familiar with linear algebra, a basisfor the space of polynomials3 is a set of polynomials such that each polynomial,whether in that set or not, can be expressed in one and only one way as a sum ofnumerical multiples of polynomials in the set.

Problem 153.◦ Show that every power of x + 1 is expressible as a sum ofnumerical multiples of powers of x. Now show that every power of x (andthus every polynomial in x) is a sum of numerical multiples (some of whichcould be negative) of powers of x + 1. This means that the powers of x + 1are a basis for the space of polynomials as well. Describe the change of basis

3The space of polynomials is just another name for the set of all polynomials.


coefficients that we use to express the binomial powers (x + 1)n in termsof the ordinary x j explicitly. Find the change of basis coefficients we use toexpress the ordinary powers xn in terms of the binomial powers (x + 1)k . (h)

Problem 154.⇒ · By multiplication, we can see that every falling factorial poly-nomial can be expressed as a sum of numerical multiples of powers of x. Insymbols, this means that there are numbers s(k , n) (notice that this s is lowercase, not upper case) such that we may write xk =

∑kn=0 s(k , n)xn . These

numbers s(k , n) are called Stirling Numbers of the first kind. By thinkingalgebraically about what the formula

xk = xk−1(x − k + 1) (3.1)

means, we can find a recurrence for Stirling numbers of the first kind thatgives us another triangular array of numbers called Stirling’s triangle of thefirst kind. Explain why Equation (3.1) is true and use it to derive a recurrencefor s(k , n) in terms of s(k − 1, n − 1) and s(k − 1, n). (h)

Problem 155. Write down the rows of Stirling’s triangle of the first kind fork = 0 to 6.

By definition, the Stirling numbers of the first kind are also change of basiscoefficients. The Stirling numbers of the first and second kind are change of basiscoefficients from the falling factorial powers of x to the ordinary factorial powers,and vice versa.

Problem 156.⇒ Explain why every rising factorial polynomial xk can be ex-pressed in terms of the falling factorial polynomials xn . Let b(k , n) standfor the change of basis coefficients that allow us to express xk in terms ofthe falling factorial polynomials xn ; that is, define b(k , n) by the equations

xk =k∑

n=0

b(k , n)xn .

(a) Find a recurrence for b(k , n). (h)

(b) Find a formula for b(k , n) and prove the correctness of what you sayin as many ways as you can. (h)

(c) Is b(k , n) the same as any of the other families of numbers (binomialcoefficients, Bell numbers, Stirling numbers, Lah numbers, etc.) wehave studied?

3.3. Partitions of Integers 63

(d) Say as much as you can (but say it precisely) about the change of basiscoefficients for expressing xk in terms of xn . (h)

3.3 Partitions of IntegersWe have now completed all our distribution problems except for those in whichboth the objects and the recipients are identical. For example, we might be puttingidentical apples into identical paper bags. In this case all that matters is how manybags get one apple (how many recipients get one object), how many get two, howmany get three, and so on. Thus for each bag we have a number, and the multiset ofnumbers of apples in the various bags is what determines our distribution of applesinto identical bags. A multiset of positive integers that add to n is called a partitionof n. Thus the partitions of 3 are 1+1+1, 1+2 (which is the same as 2+1) and 3. Thenumber of partitions of k is denoted by P(k); in computing the partitions of 3 weshowed that P(3) = 3. It is traditional to use Greek letters like λ (the Greek letterλ is pronounced LAMB duh) to stand for partitons; we might write λ = 1, 1, 1,γ = 2, 1 and τ = 3 to stand for the three partitions we just described. We also writeλ = 13 as a shorthand for λ = 1, 1, 1, and we write λ ⊣ 3 as a shorthand for “λ is apartition of three."

Problem 157.◦ Find all partitions of 4 and find all partitions of 5, therebycomputing P(4) and P(5).

3.3.1 The number of partitions of k into n partsA partition of the integer k into n parts is a multiset of n positive integers thatadd to k. We use P(k , n) to denote the number of partitions of k into n parts.Thus P(k , n) is the number of ways to distribute k identical objects to n identicalrecipients so that each gets at least one.

Problem 158.◦ Find P(6, 3) by finding all partitions of 6 into 3 parts. Whatdoes this say about the number of ways to put six identical apples into threeidentical bags so that each bag has at least one apple?

3.3.2 Representations of partitions

Problem 159.◦ How many solutions are there in the positive integers to theequation x1 + x2 + x3 = 7 with x1 ≥ x2 ≥ x3?


Problem 160. Explain the relationship between partitions of k into n partsand lists x1 , x2,. . . , xn of positive integers that add to k with x1 ≥ x2 ≥. . . ≥ xn . Such a representation of a partition is called a decreasing listrepresentation of the partition.

Problem 161.◦ Describe the relationship between partitions of k and lists orvectors (x1 , x2 , . . . , xn) such that x1+2x2+. . . kxk = k. Such a representationof a partition is called a type vector representation of a partition, and it istypical to leave the trailing zeros out of such a representation; for example(2, 1) stands for the same partition as (2, 1, 0, 0). What is the decreasing listrepresentation for this partition, and what number does it partition?

Problem 162. How does the number of partitions of k relate to the numberof partitions of k + 1 whose smallest part is one? (h)

When we write a partition as λ = λ1 , λ2 , . . . , λn , it is customary to write the listof λis as a decreasing list. When we have a type vector (t1 , t2 , . . . , tm) for a partition,we write either λ = 1t12t2 · · · mtm or λ = mtm (m − 1)tm−1 · · · 2t21t1 . Henceforth wewill use the second of these. When we write λ = λi1

1 λi22 · · · λin

n , we will assume thatλi > λi+1.

3.3.3 Ferrers and Young Diagrams and the conjugate of a partitionThe decreasing list representation of partitions leads us to a handy way to visualizepartitions. Given a decreasing list (λ1 , λ2 , . . . λn), we draw a figure made up ofrows of dots that has λ1 equally spaced dots in the first row, λ2 equally spaced dotsin the second row, starting out right below the beginning of the first row and so on.Equivalently, instead of dots, we may use identical squares, drawn so that a squaretouches each one to its immediate right or immediately below it along an edge.See Figure 3.3.1 for examples. The figure we draw with dots is called the Ferrersdiagram of the partition; sometimes the figure with squares is also called a Ferrersdiagram; sometimes it is called a Young diagram. At this stage it is irrelevantwhich name we choose and which kind of figure we draw; in more advanced workthe squares are handy because we can put things like numbers or variables intothem. From now on we will use squares and call the diagrams Young diagrams.

Figure 3.3.1: The Ferrers and Young diagrams of the partition (5,3,3,2)


Problem 163.• Draw the Young diagram of the partition (4,4,3,1,1). Describethe geometric relationship between the Young diagram of (5,3,3,2) and theYoung diagram of (4,4,3,1,1). (h)

Problem 164.• The partition (λ1 , λ2 , . . . , λn) is called the conjugate of thepartition (γ1 , γ2 , . . . , γm) if we obtain the Young diagram of one from theYoung diagram of the other by flipping one around the line with slope -1that extends the diagonal of the top left square. See Figure 3.3.2 for anexample.

Figure 3.3.2: The Ferrers diagram the partition (5,3,3,2) and its conjugate.

What is the conjugate of (4,4,3,1,1)? How is the largest part of a partitionrelated to the number of parts of its conjugate? What does this tell youabout the number of partitions of a positive integer k with largest part m? (h)

Problem 165.⇒ A partition is called self-conjugate if it is equal to its conju-gate. Find a relationship between the number of self-conjugate partitions ofk and the number of partitions of k into distinct odd parts. (h)

Problem 166. Explain the relationship between the number of partitionsof k into even parts and the number of partitions of k into parts of evenmultiplicity, i.e. parts which are each used an even number of times as in(3,3,3,3,2,2,1,1). (h)

Problem 167.⇒ Show that the number of partitions of k into four parts equalsthe number of partitions of 3k into four parts of size at most k − 1 (or 3k − 4into four parts of size at most k − 2 or 3k − 4 into four parts of size at mostk). (h)


Problem 168. The idea of conjugation of a partition could be defined with-out the geometric interpretation of a Young diagram, but it would seem farless natural without the geometric interpretation. Another idea that seemsmuch more natural in a geometric context is this. Suppose we have a par-tition of k into n parts with largest part m. Then the Young diagram of thepartition can fit into a rectangle that is m or more units wide (horizontally)and n or more units deep. Suppose we place the Young diagram of ourpartition in the top left-hand corner of an m′ unit wide and n′ unit deeprectangle with m′ ≥ m and n′ ≥ n, as in Figure 3.3.3.

Figure 3.3.3: To complement the partition (5, 3, 3, 2) in a 6 by 5 rectangle:enclose it in the rectangle, rotate, and cut out the original Young diagram.

(a) Why can we interpret the part of the rectangle not occupied by ourYoung diagram, rotated in the plane, as the Young diagram of an-other partition? This is called the complement of our partition in therectangle.

(b) What integer is being partitioned by the complement?

(c) What conditions on m′ and n′ guarantee that the complement has thesame number of parts as the original one? (h)

(d) What conditions on m′ and n′ guarantee that the complement has thesame largest part as the original one? (h)

(e) Is it possible for the complement to have both the same number ofparts and the same largest part as the original one?

(f) If we complement a partition in an m′ by n′ box and then complementthat partition in an m′ by n′ box again, do we get the same partitionthat we started with?

Problem 169.⇒ Suppose we take a partition of k into n parts with largest partm, complement it in the smallest rectangle it will fit into, complement theresult in the smallest rectangle it will fit into, and continue the process untilwe get the partition 1 of one into one part. What can you say about thepartition with which we started? (h)

Problem 170. Show that P(k , n) is at least 1n! (

k−1n−1 ). (h)


With the binomial coefficients, with Stirling numbers of the second kind, andwith the Lah numbers, we were able to find a recurrence by asking what happensto our subset, partition, or broken permutation of a set S of numbers if we removethe largest element of S. Thus it is natural to look for a recurrence to count thenumber of partitions of k into n parts by doing something similar. Unfortunately,since we are counting distributions in which all the objects are identical, there is noway for us to identify a largest element. However if we think geometrically, we canask what we could remove from a Young diagram to get a Young diagram. Twonatural ways to get a partition of a smaller integer from a partition of n would beto remove the top row of the Young diagram of the partition and to remove the leftcolumn of the Young diagram of the partition. These two operations correspondto removing the largest part from the partition and to subtracting 1 from eachpart of the partition respectively. Even though they are symmetric with respect toconjugation, they aren’t symmetric with respect to the number of parts. Thus onemight be much more useful than the other for finding a recurrence for the numberof partitions of k into n parts.

Problem 171.⇒ · In this problem we will study the two operations and seewhich one seems more useful for getting a recurrence for P(k , n).

(a) How many parts does the remaining partition have when we removethe largest part (more precisely, we reduce its multiplicity by one) froma partition of k into n parts? What can you say about the number ofparts of the remaining partition if we remove one from each part? (h)

(b) If we remove the largest part from a partition, what can we say aboutthe integer that is being partitioned by the remaining parts of thepartition? If we remove one from each part of a partition of k inton parts, what integer is being partitioned by the remaining parts?(Another way to describe this is that we remove the first column fromthe Young diagram of the partition.) (h)

(c) The last two questions are designed to get you thinking about how wecan get a bĳection between the set of partitions of k into n parts andsome other set of partitions that are partitions of a smaller number.These questions describe two different strategies for getting that set ofpartitions of a smaller number or of smaller numbers. Each strategyleads to a bĳection between partitions of k into n parts and a set ofpartitions of a smaller number or numbers. For each strategy, usethe answers to the last two questions to find and describe this set ofpartitions into a smaller number and a bĳection between partitionsof k into n parts and partitions of the smaller integer or integers intoappropriate numbers of parts. (In one case the set of partitions andbĳection are relatively straightforward to describe and in the othercase not so easy.) (h)

(d) Find a recurrence (which need not have just two terms on the righthand side) that describes how to compute P(k , n) in terms of the


number of partitions of smaller integers into a smaller number ofparts. (h)

(e) What is P(k , 1) for a positive integer k?

(f) What is P(k , k) for a positive integer k?

(g) Use your recurrence to compute a table with the values of P(k , n) forvalues of k between 1 and 7.

(h) What would you want to fill into row 0 and column 0 of your table inorder to make it consistent with your recurrence. What does this sayP(0, 0) should be? We usually define a sum with no terms in it to bezero. Is that consistent with the way the recurrence says we shoulddefine P(0, 0)? (h)

It is remarkable that there is no known formula for P(k , n), nor is there one forP(k). This section was are devoted to developing methods for computing values ofP(n , k) and finding properties of P(n , k) that we can prove even without knowinga formula. Some future sections will attempt to develop other methods.

We have seen that the number of partitions of k into n parts is equal to thenumber of ways to distribute k identical objects to n recipients so that each receivesat least one. If we relax the condition that each recipient receives at least one,then we see that the number of distributions of k identical objects to n recipientsis

∑ni=1 P(k , i) because if some recipients receive nothing, it does not matter which

recipients these are. This completes rows 7 and 8 of our table of distributionproblems. The completed table is shown in Table 3.3.4. There are quite a fewtheorems that you have proved which are summarized by Table 3.3.4. It would beworthwhile to try to write them all down!


The Twentyfold Way: A Table of Distribution Problemsk objects and conditionson how they are received

n recipients and mathematicalmodel for distribution

Distinct Identical1. Distinctno conditions

nk

functions

∑ki=1 S(n , i)

set partitions (≤ n parts)

2. DistinctEach gets at most one

nk

k-elementpermutations

1 if k ≤ n;0 otherwise

3. DistinctEach gets at least one

S(k , n)n!onto functions

S(k , n)set partitions (n parts)

4. DistinctEach gets exactly one

k! = n!permutations


5. Distinct,order matters

(k + n − 1)k

ordered functions

∑ni=1 L(k , i)

broken permutations(≤ n parts)

6. Distinct,order mattersEach gets at least one

(k)n(k − 1)k−n

orderedonto functions

L(k , n) = (kn )(k − 1)k−n

broken permutations(n parts)

7. Identicalno conditions

(n+k−1k )

multisets

∑ni=1 P(k , i)

number partitions(≤ n parts)

8. IdenticalEach gets at most one

(nk )

subsets1 if k ≤ n;

0 otherwise9. IdenticalEach gets at least one

(k−1n−1 )

compositions(n parts)

P(k , n)number partitions

(n parts)10. IdenticalEach gets exactly one



Table 3.3.4: The number of ways to distribute k objects to n recipients, with restric-tions on how the objects are received

3.3.4 Partitions into distinct partsOften Q(k , n) is used to denote the number of partitions of k into distinct parts,that is, parts that are different from each other.

Problem 172. Show that

Q(k , n) ≤ 1

n!

(k − 1

n − 1

).

(h)


Problem 173.⇒ Show that the number of partitions of 7 into 3 parts equalsthe number of partitions of 10 into three distinct parts. (h)

Problem 174.⇒ · There is a relationship between P(k , n) and Q(m , n) for someother number m. Find the number m that gives you the nicest possiblerelationship. (h)

Problem 175.· Find a recurrence that expresses Q(k , n) as a sum of Q(k −n ,m) for appropriate values of m. (h)

Problem 176.⇒ ∗ Show that the number of partitions of k into distinct partsequals the number of partitions of k into odd parts. (h)

Problem 177.⇒ ∗ Euler showed that if k " 3 j2+ j2 , then the number of partitions

of k into an even number of distinct parts is the same as the number ofpartitions of k into an odd number of distinct parts. Prove this, and in theexceptional case find out how the two numbers relate to each other. (h)

3.4 Supplementary Problems1. Answer each of the following questions with nk , kn , n!, k!, (n

k ), (kn ), nk , kn , nk ,

kn , (n+k−1k ), (n+k−1

n ), (n−1k−1 ), (

k−1n−1 ), or “none of the above".

(a) In how many ways may we pass out k identical pieces of candy to n children?(b) In how many ways may we pass out k distinct pieces of candy to n children?(c) In how many ways may we pass out k identical pieces of candy to n children

so that each gets at most one? (Assume k ≤ n.)(d) In how many ways may we pass out k distinct pieces of candy to n children

so that each gets at most one? (Assume k ≤ n.)(e) In how many ways may we pass out k distinct pieces of candy to n children

so that each gets at least one? (Assume k ≥ n.)(f) In how many ways may we pass out k identical pieces of candy to n children

so that each gets at least one? (Assume k ≥ n.)

2. The neighborhood betterment committee has been given r trees to distribute tos families living along one side of a street.


(a) In how many ways can they distribute all of them if the trees are distinct, thereare more families than trees, and each family can get at most one?

(b) In how many ways can they distribute all of them if the trees are distinct, anyfamily can get any number, and a family may plant its trees where it chooses?

(c) In how many ways can they distribute all the trees if the trees are identical,there are no more trees than families, and any family receives at most one?

(d) In how many ways can they distribute them if the trees are distinct, there aremore trees than families, and each family receives at most one (so there couldbe some leftover trees)?

(e) In how many ways can they distribute all the trees if they are identical andanyone may receive any number of trees?

(f) In how many ways can all the trees be distributed and planted if the trees aredistinct, any family can get any number, and a family must plant its trees inan evenly spaced row along the road?

(g) Answer the question in Part 3.4.2.f assuming that every family must get a tree.(h) Answer the question in Part 3.4.2.e assuming that each family must get at least

one tree.

3. In how many ways can n identical chemistry books, r identical mathematicsbooks, s identical physics books, and t identical astronomy books be arranged onthree bookshelves? (Assume there is no limit on the number of books per shelf.)

4.⇒ One formula for the Lah numbers is

L(k , n) =(kn

)(k − 1)k−n

Find a proof that explains this product.

5. What is the number of partitions of n into two parts?

6. What is the number of partitions of k into k − 2 parts?

7. Show that the number of partitions of k into n parts of size at most m equals thenumber of partitions of mn − k into no more than n parts of size at most m − 1.

8. Show that the number of partitions of k into parts of size at most m is equal tothe number of partitions of of k + m into m parts.

9. You can say something pretty specific about self-conjugate partitions of k intodistinct parts. Figure out what it is and prove it. With that, you should be able to finda relationship between these partitions and partitions whose parts are consecutiveintegers, starting with 1. What is that relationship?

10. What is s(k , 1)?

11. Show that the Stirling numbers of the second kind satisfy the recurrence

S(k , n) =k∑

i=1

S(k − i , n − 1)

(n − 1

i − 1

).


12.⇒ Let c(k , n) be the number of ways for k children to hold hands to form n circles,where one child clasping his or her hands together and holding them out to form acircle is considered a circle. Find a recurrence for c(k , n). Is the family of numbersc(k , n) related to any of the other families of numbers we have studied? If so, how?

13.⇒ How many labeled trees on n vertices have exactly four vertices of degree 1?

14.⇒ The degree sequence of a graph is a list of the degrees of the vertices in non-increasing order. For example the degree sequence of the first graph in Figure 2.3.3is (4, 3, 2, 2, 1). For a graph with vertices labeled 1 through n, the ordered degreesequence of the graph is the sequence d1 , d2 , . . . , dn in which di is the degree ofvertex i. For example the ordered degree sequence of the first graph in Figure 2.3.1is (1, 2, 3, 3, 1, 1, 2, 1).

(a) How many labeled trees are there on n vertices with ordered degree sequenced1 , d2 , . . . , dn?

(b) How many labeled trees are there on n vertices with with the degree sequencein which the degree d appears id times?

Chapter 4

Generating Functions

4.1 The Idea of Generating Functions4.1.1 Visualizing Counting with PicturesSuppose you are going to choose three pieces of fruit from among apples, pearsand bananas for a snack. We can symbolically represent all your choices as

🍎🍎🍎+🍐🍐🍐+🍌🍌🍌+🍎🍎🍐+🍎🍎🍌+🍎🍐🍐+🍐🍐🍌+🍎🍌🍌+🍐🍌🍌+🍎🍐🍌.

Here we are using a picture of a piece of fruit to stand for taking a piece of that fruit.Thus 🍎 stands for taking an apple, 🍎🍐 for taking an apple and a pear, and 🍎🍎for taking two apples. You can think of the plus sign as standing for the “exclusiveor,” that is,🍎+🍌 would stand for “I take an apple or a banana but not both.” Tosay “I take both an apple and a banana,” we would write🍎🍌. We can extend theanalogy to mathematical notation by condensing our statement that we take threepieces of fruit to

🍎3 + 🍐3 +🍌3 +🍎2🍐+🍎2🍌+🍎🍐2 + 🍐2🍌+🍎🍌2 + 🍐🍌2 +🍎🍐🍌.

In this notation 🍎3 stands for taking a multiset of three apples, while 🍎2🍌stands for taking a multiset of two apples and a banana, and so on. What ournotation is really doing is giving us a convenient way to list all three elementmultisets chosen from the set {🍎,🍐,🍌}.1

Suppose now that we plan to choose between one and three apples, betweenone and two pears, and between one and two bananas. In a somewhat clumsy waywe could describe our fruit selections as

🍎🍐🍌+🍎2🍐🍌 + · · ·+🍎2🍐2🍌 + · · ·+🍎2🍐2🍌2

+🍎3🍐🍌 + · · ·+🍎3🍐2🍌 + · · ·+🍎3🍐2🍌2. (4.1)

1This approach was inspired by George Pólya’s paper “Picture Writing,” in the December, 1956 issueof the American Mathematical Monthly, page 689. While we are taking a somewhat more formal approachthan Pólya, it is still completely in the spirit of his work.

73

74 4. Generating Functions

Problem 178.• Using an A in place of the picture of an apple, a P in placeof the picture of a pear, and a B in place of the picture of a banana, writeout the formula similar to Formula (4.1) without any dots for left out terms.(You may use pictures instead of letters if you prefer, but it gets tediousquite quickly!) Now expand the product (A + A2 + A3)(P + P2)(B + B2)and compare the result with your formula.

Problem 179.• Substitute x for all of A, P and B (or for the correspondingpictures) in the formula you got in Problem 178 and expand the result inpowers of x. Give an interpretation of the coefficient of xn .

If we were to expand the formula

(🍎+🍎2 +🍎3)(🍐+ 🍐2)(🍌+🍌2). (4.2)

we would get Formula (4.1). Thus Formula (4.1) and Formula (4.2) each describethe number of multisets we can choose from the set {🍎,🍐,🍌} in which🍎 appearsbetween 1 and three times and 🍐 and 🍌 each appear once or twice. We interpretFormula (4.1) as describing each individual multiset we can choose, and we inter-pret Formula (4.2) as saying that we first decide how many apples to take, and thendecide how many pears to take, and then decide how many bananas to take. Atthis stage it might seem a bit magical that doing ordinary algebra with the secondformula yields the first, but in fact we could define addition and multiplicationwith these pictures more formally so we could explain in detail why things workout. However since the pictures are for motivation, and are actually difficult towrite out on paper, it doesn’t make much sense to work out these details. We willsee an explanation in another context later on.

4.1.2 Picture functionsAs you’ve seen, in our descriptions of ways of choosing fruits, we’ve treated thepictures of the fruit as if they are variables. You’ve also likely noticed that it ismuch easier to do algebraic manipulations with letters rather than pictures, simplybecause it is time consuming to draw the same picture over and over again, whilewe are used to writing letters quickly. In the theory of generating functions,we associate variables or polynomials or even power series with members of a set.There is no standard language describing how we associate variables with membersof a set, so we shall invent2 some. By a picture of a member of a set we will meana variable, or perhaps a product of powers of variables (or even a sum of productsof powers of variables). A function that assigns a picture P(s) to each member sof a set S will be called a picture function . The picture enumerator for a picturefunction P defined on a set S will be

EP(S) =∑s:s∈S

P(s).

2We are really adapting language introduced by George Pólya.

4.1. The Idea of Generating Functions 75

We choose this language because the picture enumerator lists, or enumerates,all the elements of S according to their pictures. Thus Formula (4.1) is the pictureenumerator the set of all multisets of fruit with between one and three apples, oneand two pears, and one and two bananas.

Problem 180.◦ How would you write down a polynomial in the variable Athat says you should take between zero and three apples?

Problem 181.• How would you write down a picture enumerator that sayswe take between zero and three apples, between zero and three pears, andbetween zero and three bananas?

Problem 182.· (Used in Chapter 6.) Notice that when we used A2 to stand fortaking two apples, and P3 to stand for taking three pears, then we used theproduct A2P3 to stand for taking two apples and three pears. Thus we havechosen the picture of the ordered pair (2 apples, 3 pears) to be the productof the pictures of a multiset of two apples and a multiset of three pears.Show that if S1 and S2 are sets with picture functions P1 and P2 definedon them, and if we define the picture of an ordered pair (x1 , x2) ∈ S1 × S2

to be P((x1 , x2)) = P1(x1)P2(x2), then the picture enumerator of P on theset S1 × S2 is EP1(S1)EP2(S2). We call this the product principle for pictureenumerators.

4.1.3 Generating functions

Problem 183.• Suppose you are going to choose a snack of between zeroand three apples, between zero and three pears, and between zero andthree bananas. Write down a polynomial in one variable x such that thecoefficient of xn is the number of ways to choose a snack with n pieces offruit. (h)

Problem 184.◦ Suppose an apple costs 20 cents, a banana costs 25 cents,and a pear costs 30 cents. What should you substitute for A, P, and B inProblem 181 in order to get a polynomial in which the coefficient of xn isthe number of ways to choose a selection of fruit that costs n cents? (h)


Problem 185.• Suppose an apple has 40 calories, a pear has 60 calories, anda banana has 80 calories. What should you substitute for A, P, and B inProblem 181 in order to get a polynomial in which the coefficient of xn isthe number of ways to choose a selection of fruit with a total of n calories?

Problem 186.• We are going to choose a subset of the set [n] = {1, 2, . . . , n}.Suppose we use x1 to be the picture of choosing 1 to be in our subset. Whatis the picture enumerator for either choosing 1 or not choosing 1? Supposethat for each i between 1 and n, we use xi to be the picture of choosing ito be in our subset. What is the picture enumerator for either choosing i ornot choosing i to be in our subset? What is the picture enumerator for allpossible choices of subsets of [n]? What should we substitute for xi in orderto get a polynomial in x such that the coefficient of xk is the number of waysto choose a k-element subset of n? What theorem have we just reproved (aspecial case of)? (h)

In Problem 186 we see that we can think of the process of expanding the polyno-mial (1+x)n as a way of “generating” the binomial coefficients (n

k ) as the coefficientsof xk in the expansion of (1 + x)n . For this reason, we say that (1 + x)n is the “gen-erating function” for the binomial coefficients (n

k ). More generally, the generatingfunction for a sequence ai , defined for i with 0 ≤ i ≤ n is the expression

∑ni=0 ai xi ,

and the generating function for the sequence ai with i ≥ 0 is the expression∑∞i=0 ai xi . This last expression is an example of a power series. In calculus it is

important to think about whether a power series converges in order to determinewhether or not it represents a function. In a nice twist of language, even though weuse the phrase generating function as the name of a power series in combinatorics,we don’t require the power series to actually represent a function in the usual sense,and so we don’t have to worry about convergence.3 Instead we think of a powerseries as a convenient way of representing the terms of a sequence of numbers ofinterest to us. The only justification for saying that such a representation is con-venient is because of the way algebraic properties of power series capture some ofthe important properties of some sequences that are of combinatorial importance.The remainder of this chapter is devoted to giving examples of how the algebra ofpower series reflects combinatorial ideas.

Because we choose to think of power series as strings of symbols that we ma-nipulate by using the ordinary rules of algebra and we choose to ignore issuesof convergence, we have to avoid manipulating power series in a way that wouldrequire us to add infinitely many real numbers. For example, we cannot make thesubstitution of y +1 for x in the power series

∑∞i=0 xi , because in order to interpret∑∞

i=0(y + 1)i as a power series we would have to apply the binomial theorem toeach of the (y + 1)i terms, and then collect like terms, giving us infinitely many

3In the evolution of our current mathematical terminology, the word function evolved throughseveral meanings, starting with very imprecise meanings and ending with our current rather precisemeaning. The terminology “generating function” may be thought of as an example of one of the earlierusages of the term function.


ones added together as the coefficient of y0, and in fact infinitely many numbersadded together for the coefficient of any yi . (On the other hand, it would be fine tosubstitute y + y2 for x. Can you see why?)

4.1.4 Power seriesFor now, most of our uses of power series will involve just simple algebra. Sincewe use power series in a different way in combinatorics than we do in calculus, weshould review a bit of the algebra of power series.

Problem 187.• In the polynomial (a0 + a1x + a2x2)(b0 + b1x + b2x2 + b3x3),what is the coefficient of x2? What is the coefficient of x4?

Problem 188.• In Problem 187 why is there a b0 and a b1 in your expressionfor the coefficient of x2 but there is not a b0 or a b1 in your expression forthe coefficient of x4? What is the coefficient of x4 in

(a0 + a1x + a2x2 + a3x3 + a4x4)(b0 + b1x + b2x2 + b3x3 + b4x4)?

Express this coefficient in the form

4∑i=0

something,

where the something is an expression you need to figure out. Now supposethat a3 = 0, a4 = 0 and b4 = 0. To what is your expression equal afteryou substitute these values? In particular, what does this have to do withProblem 187? (h)

Problem 189.• The point of the Problems 187 and Problem 188 is that so longas we are willing to assume ai = 0 for i > n and b j = 0 for j > m, then thereis a very nice formula for the coefficient of xk in the product

(n∑

i=0

ai xi

) -./

m∑j=0

b j x j012.

Write down this formula explicitly. (h)

Problem 190.• Assuming that the rules you use to do arithmetic with poly-nomials apply to power series, write down a formula for the coefficient of


xk in the product ( ∞∑i=0

ai xi

) -./

∞∑j=0

b j x j012

.

(h)

We use the expression you obtained in Problem 190 to define the product ofpower series. That is, we define the product

( ∞∑i=0

ai xi

) -./

∞∑j=0

b j x j012

to be the power series∑∞

k=0 ck xk , where ck is the expression you found in Prob-lem 190. Since you derived this expression by using the usual rules of algebra forpolynomials, it should not be surprising that the product of power series satisfiesthese rules.4

4.1.5 Product principle for generating functionsEach time that we converted a picture function to a generating function by substi-tuting x or some power of x for each picture, the coefficient of x had a meaningthat was significant to us. For example, with the picture enumerator for selectingbetween zero and three each of apples, pears, and bananas, when we substitutedx for each of our pictures, the exponent i in the power xi is the number of piecesof fruit in the fruit selection that led us to xi . After we simplify our product bycollecting together all like powers of x, the coefficient of xi is the number of fruitselections that use i pieces of fruit. In the same way, if we substitute xc for a pic-ture, where c is the number of calories in that particular kind of fruit, then the i inan xi term in our generating function stands for the number of calories in a fruitselection that gave rise to xi , and the coefficient of xi in our generating functionis the number of fruit selections with i calories. The product principle of pictureenumerators translates directly into a product principle for generating functions.

Problem 191.• Suppose that we have two sets S1 and S2. Let v1 (v standsfor value) be a function from S1 to the nonnegative integers and let v2 be afunction from S2 to the nonnegative integers. Define a new function v onthe set S1×S2 by v(x1 , x2) = v1(x1)+v2(x2). Suppose further that

∑∞i=0 ai xi

is the generating function for the number of elements x1 of S1 of value i, thatis with v1(x1) = i. Suppose also that

∑∞j=0 b j x j is the generating function

for the number of elements x2 of S2 of value j, that is with v2(x2) = j. Prove

4Technically we should explicitly state these rules and prove that they are all valid for power seriesmultiplication, but it seems like overkill at this point to do so!


that the coefficient of xk in( ∞∑

i=0

ai xi

) -./

∞∑j=0

b j x j012

is the number of ordered pairs (x1 , x2) in S1 × S2 with total value k, that iswith v1(x1)+v2(x2) = k. This is called the product principle for generatingfunctions. (h)

Problem 191 may be extended by mathematical induction to prove our nexttheorem.

Theorem 4.1.1. If S1 , S2 , . . . , Sn are sets with a value function vi from Si to the nonneg-ative integers for each i and fi(x) is the generating function for the number of elements ofSi of each possible value, then the generating function for the number of n-tuples of eachpossible value is

∏ni=1 fi(x).

4.1.6 The extended binomial theorem and multisets

Problem 192.• Suppose once again that i is an integer between 1 and n.

(a) What is the generating function in which the coefficient of xk is 1? Thisseries is an example of what is called an infinite geometric series. Inthe next part of this problem, it will be useful to interpret the coefficientone as the number of multisets of size k chosen from the singleton set{i}. Namely, there is only one way to choose a multiset of size k from{i}: choose i exactly k times.

(b) Express the generating function in which the coefficient of xk is thenumber of multisets chosen from [n] as a power of a power series.What does Problem 125 (in which your answer could be expressedas a binomial coefficient) tell you about what this generating functionequals? (h)

Problem 193.◦ What is the product (1 − x)∑n

k=0 xk? What is the product

(1 − x)∞∑

k=0

xk?

Problem 194. Express the generating function for the number of multisetsof size k chosen from [n] (where n is fixed but k can be any nonnegativeinteger) as a 1 over something relatively simple.


Problem 195.• Find a formula for (1 + x)−n as a power series whose coeffi-cients involve binomial coefficients. What does this formula tell you abouthow we should define (−n

k ) when n is positive? (h)

Problem 196. If you define (−nk ) in the way you described in Problem 195,

you can write down a version of the binomial theorem for (x + y)n that isvalid for both nonnegative and negative values of n. Do so. This is calledthe extended binomial theorem. Write down a special case with n negative,like n = −3, to see an interesting surprise that suggests why we do not usethis formula later on.

Problem 197. Write down the generating function for the number of waysto distribute identical pieces of candy to three children so that no childgets more than 4 pieces. Write this generating function as a quotient ofpolynomials. Using both the extended binomial theorem and the originalbinomial theorem, find out in how many ways we can pass out exactly tenpieces. (h)

Problem 198.• What is the generating function for the number of multisetschosen from an n-element set so that each element appears at least j timesand less than m times? Write this generating function as a quotient ofpolynomials, then as a product of a polynomial and a power series. (h)

Problem 199.⇒ Recall that a tree is determined by its edge set. Suppose youhave a tree on n vertices, say with vertex set [n]. We can use xi as the pictureof vertex i and xi x j as the picture of the edge xi x j . Then one possible pictureof the tree T is the product P(T) =

∏{i , j}:i and j are adjacent xi x j .

(a) Explain why the picture of a tree is also∏n

i=1 x (i)i .

(b) Write down the picture enumerators for trees on two, three, and fourvertices. Factor them as completely as possible.

(c) Explain why x1x2 · · · xn is a factor of the picture of a tree on n vertices.

(d) Write down the picture of a tree on five vertices with one vertex ofdegree four, say vertex i. If a tree on five vertices has a vertex of degreethree, what are the possible degrees of the other vertices. What canyou say about the picture of a tree with a vertex of degree three? Ifa tree on five vertices has no vertices of degree three or four, how

4.2. Generating functions for integer partitions 81

many vertices of degree two does it have? What can you say about itspicture? Write down the picture enumerator for trees on five vertices.

(e) Find a (relatively) simple polynomial expression for the picture enu-merator

∑T : T is a tree on [n] P(T). Prove it is correct. (h)

(f) The enumerator for trees by degree sequence is the sum over all treesof xd1 xd2 · · · xdn , where di is the degree of vertex i. What is the enu-merator by degree sequence for trees on the vertex set [n]?

4.2 Generating functions for integer partitions

Problem 200.• If we have five identical pennies, five identical nickels, fiveidentical dimes, and five identical quarters, give the picture enumeratorfor the combinations of coins we can form and convert it to a generatingfunction for the number of ways to make k cents with the coins we have. Dothe same thing assuming we have an unlimited supply of pennies, nickels,dimes, and quarters. (h)

Problem 201.• Recall that a partition of an integer k is a multiset of numbersthat adds to k. In Problem 200 we found the generating function for thenumber of partitions of an integer into parts of size 1, 5, 10, and 25. Whenworking with generating functions for partitions, it is becoming standard touse q rather than x as the variable in the generating function. Write youranswers in this notation.a

(a) Give the generating function for the number partitions of an integerinto parts of size one through ten. (h)

(b) Give the generating function for the number of partitions of an integerk into parts of size at most m, where m is fixed but k may vary. Noticethis is the generating function for partitions whose Young diagramfits into the space between the line x = 0 and the line x = m in acoordinate plane. (We assume the boxes in the Young diagram areone unit by one unit.) (h)

aThe reason for this change in the notation relates to the subject of finite fields in abstractalgebra, where q is the standard notation for the size of a finite field. While we will make nouse of this connection, it will be easier for you to read more advanced work if you get used tothe different notation.


Problem 202.• In Problem 201.b you gave the generating function for thenumber of partitions of an integer into parts of size at most m. Explainwhy this is also the generating function for partitions of an integer into atmost m parts. Notice that this is the generating function for the number ofpartitions whose Young diagram fits into the space between the line y = 0and the line y = m. (h)

Problem 203.• When studying partitions of integers, it is inconvenient torestrict ourselves to partitions with at most m parts or partitions with max-imum part size m.

(a) Give the generating function for the number of partitions of an integerinto parts of any size. Don’t forget to use q rather than x as yourvariable. (h)

(b) Find the coefficient of q4 in this generating function. (h)

(c) find the coefficient of q5 in this generating function.

(d) This generating function involves an infinite product. Describe theprocess you would use to expand this product into as many terms ofa power series as you choose. (h)

(e) Rewrite any power series that appear in your product as quotients ofpolynomials or as integers divided by polynomials.

Problem 204.⇒ In Problem 203, we multiplied together infinitely many powerseries. Here are two notations for infinite products that look rather similar:

∞∏i=1

1 + x + x2 + · · ·+ xi and∞∏

i=1

1 + xi + x2i + · · ·+ xi2 .

However, one makes sense and one doesn’t. Figure out which one makessense and explain why it makes sense and the other one doesn’t. If we wanta product of the form

∞∏i=1

1 + pi(x),

where each pi(x) is a nonzero polynomial in x to make sense, describe arelatively simple assumption about the polynomials pi(x) that will make theproduct make sense. If we assumed the terms pi(x) were nonzero powerseries, is there a relatively simple assumption we could make about themin order to make the product make sense? (Describe such a condition orexplain why you think there couldn’t be one.) (h)

4.2. Generating functions for integer partitions 83

Problem 205.• What is the generating function (using q for the variable) forthe number of partitions of an integer in which each part is even? (h)

Problem 206.• What is the generating function (using q as the variable) forthe number of partitions of an integer into distinct parts, that is, in whicheach part is used at most once? (h)

Problem 207.• Use generating functions to explain why the number of par-titions of an integer in which each part is used an even number of timesequals the generating function for the number of partitions of an integer inwhich each part is even. (h)

Problem 208. Use the fact that

1 − q2i

1 − qi = 1 + qi

and the generating function for the number of partitions of an integer intodistinct parts to show how the number of partitions of an integer k intodistinct parts is related to the number of partitions of an integer k into oddparts. (h)

Problem 209. Write down the generating function for the number of waysto partition an integer into parts of size no more than m, each used anodd number of times. Write down the generating function for the numberof partitions of an integer into parts of size no more than m, each usedan even number of times. Use these two generating functions to get arelationship between the two sequences for which you wrote down thegenerating functions. (h)

Problem 210.⇒ In Problem 201.b and Problem 202 you gave the generatingfunctions for, respectively, the number of partitions of k into parts the largestof which is at most m and for the number of partitions of k into at most mparts. In this problem we will give the generating function for the numberof partitions of k into at most n parts, the largest of which is at most m.That is we will analyze

∑∞i=0 ak qk where ak is the number of partitions of k

into at most n parts, the largest of which is at most m. Geometrically, it is


the generating function for partitions whose Young diagram fits into an mby n rectangle, as in Problem 167. This generating function has significantanalogs to the binomial coefficient (m+n

n ), and so it is denoted by [m+nn ]q . It

is called a q-binomial coefficient.

(a) Compute [42 ]q = [2+22 ]q . (h)

(b) Find explicit formulas for [n1 ]q and [ nn−1 ]q . (h)

(c) How are [m+nn ]q and [m+n

n ]q related? Prove it. (Note this is the same asasking how [rs ]q and [ r

r−s ]q are related.) (h)

(d) So far the analogy to (m+nn ) is rather thin! If we had a recurrence

like the Pascal recurrence, that would demonstrate a real analogy. Is[m+n

n ]q = [m+n−1n−1 ]q + [m+n−1

n ]q?

(e) Recall the two operations we studied in Problem 171.

(i) The largest part of a partition counted by [m+nn ]q is either m or

is less than or equal to m − 1. In the second case, the partitionfits into a rectangle that is at most m − 1 units wide and at mostn units deep. What is the generating function for partitions ofthis type? In the first case, what kind of rectangle does thepartition we get by removing the largest part sit in? What is thegenerating function for partitions that sit in this kind of rectangle?What is the generating function for partitions that sit in this kindof rectangle after we remove a largest part of size m? Whatrecurrence relation does this give you?

(ii) What recurrence do you get from the other operation we studiedin Problem 171?

(iii) It is quite likely that the two recurrences you got are different.One would expect that they might give different values for [m+n

n ]q .Can you resolve this potential conflict? (h)

(f) Define [n]q to be 1 + q + · · · + qn−1 for n > 0 and [0]q = 1. We readthis simply as n-sub-q. Define [n]!q to be [n]q [n−1]q · · · [3]q [2]q [1]q . Weread this as n cue-torial, and refer to it as a q-ary factorial. Show that

[m + n

n

]q=

[m + n]!q[m]!q [n]!q

.

(h)

(g) Now think of q as a variable that we will let approach 1. Find anexplicit formula for

(i)q→1

[n]q .

4.3. Generating Functions and Recurrence Relations 85

(ii)q→1

[n]!q .

(iii)q→1

[m + n

n

]q.

Why is the limit in Part iii equal to the number of partitions (of anynumber) with at most n parts all of size most m? Can you explainbĳectively why this quantity equals the formula you got? (h)

(h)∗ What happens to [m+nn ]q if we let q approach −1? (h)

4.3 Generating Functions and Recurrence RelationsRecall that a recurrence relation for a sequence an expresses an in terms of valuesai for i < n. For example, the equation ai = 3ai−1+2i is a first order linear constantcoefficient recurrence.

4.3.1 How generating functions are relevantAlgebraic manipulations with generating functions can sometimes reveal the solu-tions to a recurrence relation.

Problem 211.• Suppose that ai = 3ai−1 + 3i .

(a) Multiply both sides by xi and sum both the left hand side and righthand side from i = 1 to infinity. In the left-hand side use the fact that

∞∑i=1

ai xi = (∞∑

i=0

xi) − a0

and in the right hand side, use the fact that∞∑

i=1

ai−1xi = x∞∑

i=1

ai xi−1 = x∞∑

j=0

a j x j = x∞∑

i=0

ai xi

(where we substituted j for i−1 to see explicitly how to change the lim-its of summation, a surprisingly useful trick) to rewrite the equationin terms of the power series

∑∞i=0 ai xi . Solve the resulting equation

for the power series∑∞

i=0 ai xi . You can save a lot of writing by usinga variable like y to stand for the power series.

(b) Use the previous part to get a formula for ai in terms of a0.

(c) Now suppose that ai = 3ai−1 + 2i . Repeat the previous two steps forthis recurrence relation. (There is a way to do this part using what youalready know. Later on we shall introduce yet another way to dealwith the kind of generating function that arises here.) (h)


Problem 212.◦ Suppose we deposit $5000 in a savings certificate that paysten percent interest and also participate in a program to add $1000 to thecertificate at the end of each year (from the end of the first year on) thatfollows (also subject to interest.) Assuming we make the $5000 deposit atthe end of year 0, and letting ai be the amount of money in the account atthe end of year i, write a recurrence for the amount of money the certificateis worth at the end of year n. Solve this recurrence. How much money dowe have in the account (after our year-end deposit) at the end of ten years?At the end of 20 years?

4.3.2 Fibonacci numbersThe sequence of problems that follows (culminating in Problem 222) describes anumber of hypotheses we might make about a fictional population of rabbits. Weuse the example of a rabbit population for historic reasons; our goal is a classicalsequence of numbers called Fibonacci numbers. When Fibonacci5 introduced them,he did so with a fictional population of rabbits.

4.3.3 Second order linear recurrence relations

Problem 213.• Suppose we start (at the end of month 0) with 10 pairs ofbaby rabbits, and that after baby rabbits mature for one month they begin toreproduce, with each pair producing two new pairs at the end of each monthafterwards. Suppose further that over the time we observe the rabbits, nonedie. Let an be the number of rabbits we have at the end of month n. Showthat an = an−1+2an−2. This is an example of a second order linear recurrencewith constant coefficients. Using a method similar to that of Problem 211,show that ∞∑

i=0

ai xi =10

1 − x − 2x2.

This gives us the generating function for the sequence ai giving the pop-ulation in month i; shortly we shall see a method for converting this to asolution to the recurrence.

Problem 214.• In Fibonacci’s original problem, each pair of mature rabbitsproduces one new pair at the end of each month, but otherwise the situationis the same as in Problem 213. Assuming that we start with one pair of babyrabbits (at the end of month 0), find the generating function for the numberof pairs of rabbits we have at the end on n months. (h)

5Apparently Leanardo de Pisa was given the name Fibonacci posthumously. It is a shortening of“son of Bonacci” in Italian.


Problem 215.⇒ Find the generating function for the solutions to the recur-rence

ai = 5ai−1 − 6ai−2 + 2i .

The recurrence relations we have seen in this section are called second orderbecause they specify ai in terms of ai−1 and ai−2, they are called linear because ai−1and ai−2 each appear to the first power, and they are called constant coefficientrecurrences because the coefficients in front of ai−1 and ai−2 are constants.

4.3.4 Partial fractionsThe generating functions you found in the previous section all can be expressedin terms of the reciprocal of a quadratic polynomial. However without a powerseries representation, the generating function doesn’t tell us what the sequenceis. It turns out that whenever you can factor a polynomial into linear factors (andover the complex numbers such a factorization always exists) you can use thatfactorization to express the reciprocal in terms of power series.

Problem 216.• Express 1x−3 + 2

x−2 as a single fraction.

Problem 217.◦ In Problem 216 you see that when we added numerical multi-ples of the reciprocals of first degree polynomials we got a fraction in whichthe denominator is a quadratic polynomial. This will always happen unlessthe two denominators are multiples of each other, because their least com-mon multiple will simply be their product, a quadratic polynomial. Thisleads us to ask whether a fraction whose denominator is a quadratic poly-nomial can always be expressed as a sum of fractions whose denominatorsare first degree polynomials. Find numbers c and d so that

5x + 1

(x − 3)(x + 5)=

cx − 3

+d

x + 5.

(h)

Problem 218.• In Problem 217 you may have simply guessed at values of cand d, or you may have solved a system of equations in the two unknownsc and d. Given constants a, b, r1, and r2 (with r1 " r2), write down a systemof equations we can solve for c and d to write

ax + b(x − r1)(x − r2)

=c

x − r1+

dx − r2

.

(h)


Writing down the equations in Problem 218 and solving them is called themethod of partial fractions. This method will let you find power series expansionsfor generating functions of the type you found in Problems 213 to Problem 215.However you have to be able to factor the quadratic polynomials that are in thedenominators of your generating functions.

Problem 219.• Use the method of partial fractions to convert the generatingfunction of Problem 213 into the form

cx − r1

+d

x − r2.

Use this to find a formula for an .

Problem 220.• Use the quadratic formula to find the solutions to x2+x−1 =0, and use that information to factor x2 + x − 1.

Problem 221.• Use the factors you found in Problem 220 to write

1

x2 + x − 1

in the formc

x − r1+

dx − r2

.

(h)

Problem 222.•

(a) Use the partial fractions decomposition you found in Problem 220 towrite the generating function you found in Problem 214 in the form

∞∑n=0

an xi

and use this to give an explicit formula for an . (h)

(b) When we have a0 = 1 and a1 = 1, i.e. when we start with one pairof baby rabbits, the numbers an are called Fibonacci Numbers. Useeither the recurrence or your final formula to find a2 through a8. Areyou amazed that your general formula produces integers, or for thatmatter produces rational numbers? Why does the recurrence equationtell you that the Fibonacci numbers are all integers?


(c) Explain why there is a real number b such that, for large values of n,the value of the nth Fibonacci number is almost exactly (but not quite)some constant times bn . (Find b and the constant.)

(d) Find an algebraic explanation (not using the recurrence equation) ofwhat happens to make the square roots of five go away. (h)

(e) As a challenge (which the author has not yet done), see if you can finda way to show algebraically (not using the recurrence relation, butrather the formula you get by removing the square roots of five) thatthe formula for the Fibonacci numbers yields integers.

Problem 223. Solve the recurrence an = 4an−1 − 4an−2.

4.3.5 Catalan Numbers

Problem 224.⇒

(a) Using either lattice paths or diagonal lattice paths, explain why theCatalan Number cn satisfies the recurrence

Cn =n−1∑i=1

Ci−1Cn−i .

(h)

(b) Show that if we use y to stand for the power series∑∞

n=0 cn xn , thenwe can find y by solving a quadratic equation. Find y. (h)

(c) Taylor’s theorem from calculus tells us that the extended binomialtheorem

(1 + x)r =∞∑

i=0

(ri

)xi

holds for any number real number r, where (ri ) is defined to be

ri

i!=

r(r − 1) · · · (r − i + 1)

i!.

Use this and your solution for y (note that of the two possible valuesfor y that you get from the quadratic formula, only one gives an actualpower series) to get a formula for the Catalan numbers. (h)


4.4 Supplementary Problems1.⇒ ∗ What is the generating function for the number of ways to pass out k pieces ofcandy from an unlimited supply of identical candy to n children (where n is fixed)so that each child gets between three and six pieces of candy (inclusive)? Use thefact that

(1 + x + x2 + x3)(1 − x) = 1 − x4

to find a formula for the number of ways to pass out the candy.

2.◦(a) In paying off a mortgage loan with initial amount A, annual interest rate p%

on a monthly basis with a monthly payment of m, what recurrence describesthe amount owed after n months of payments in terms of the amount owedafter n − 1 months? Some technical details: You make the first payment afterone month. The amount of interest included in your monthly payment is.01p/12. This interest rate is applied to the amount you owed immediatelyafter making your last monthly payment.

(b) Find a formula for the amount owed after n months.(c) Find a formula for the number of months needed to bring the amount owed

to zero. Another technical point: If you were to make the standard monthlypayment m in the last month, you might actually end up owing a negativeamount of money. Therefore it is ok if the result of your formula for thenumber of months needed gives a non-integer number of months. The bankwould just round up to the next integer and adjust your payment so yourbalance comes out to zero.

(d) What should the monthly payment be to pay off the loan over a period of 30years?

3.⇒ We have said that for nonnegative i and positive n we want to define (−ni ) to be

(n+i−1i ). If we want the Pascal recurrence to be valid, how should we define (−n

−i )when n and i are both positive?

4.⇒ Find a recurrence relation for the number of ways to divide a convex n-gon intotriangles by means of non-intersecting diagonals. How do these numbers relate tothe Catalan numbers?

5.⇒ How does∑n

k=0 (n−k

k ) relate to the Fibonacci Numbers?

6. Let m and n be fixed. Express the generating function for the number of k-element multisets of an n-element set such that no element appears more than mtimes as a quotient of two polynomials. Use this expression to get a formula for thenumber of k-element multisets of an n-element set such that no element appearsmore than m times.

7. One natural but oversimplified model for the growth of a tree is that all newwood grows from the previous year’s growth and is proportional to it in amount.To be more precise, assume that the (total) length of the new growth in a givenyear is the constant c times the (total) length of new growth in the previous year.


Write down a recurrence for the total length an of all the branches of the tree atthe end of growing season n. Find the general solution to your recurrence relation.Assume that we begin with a one meter cutting of new wood (from the previousyear) which branches out and grows a total of two meters of new wood in the firstyear. What will the total length of all the branches of the tree be at the end of nyears?

8.⇒ (Relevant to Appendix C) We have some chairs which we are going to paint withred, white, blue, green, yellow and purple paint. Suppose that we may paint anynumber of chairs red or white, that we may paint at most one chair blue, at mostthree chairs green, only an even number of chairs yellow, and only a multiple offour chairs purple. In how many ways may we paint n chairs?

9. What is the generating function for the number of partitions of an integer inwhich each part is used at most m times? Why is this also the generating functionfor partitions in which consecutive parts (in a decreasing list representation) differby at most m and the smallest part is also at most m?

Chapter 5

The Principle of Inclusion andExclusion

5.1 The size of a union of setsOne of our very first counting principles was the sum principle which says thatthe size of a union of disjoint sets is the sum of their sizes. Computing the sizeof overlapping sets requires, quite naturally, information about how they overlap.Taking such information into account will allow us to develop a powerful extensionof the sum principle known as the “principle of inclusion and exclusion.”

5.1.1 Unions of two or three sets

Problem 225.◦ In a biology lab study of the effects of basic fertilizer ingredi-ents on plants, 16 plants are treated with potash, 16 plants are treated withphosphate, and among these plants, eight are treated with both phosphateand potash. No other treatments are used. How many plants receive at leastone treatment? If 32 plants are studied, how many receive no treatment?

Problem 226.+ Give a formula for the size of the union A ∪ B of two sets Ain terms of the sizes |A| of A, |B | of B, and |A ∩ B | of A ∩ B. If A and Bare subsets of some “universal” set U, express the size of the complementU − (A ∪ B) in terms of the sizes |U | of U, |A| of A, |B | of B, and |A ∩ B | ofA ∩ B. (h)

Problem 227.◦ In Problem 225, there were just two fertilizers used to treatthe sample plants. Now suppose there are three fertilizer treatments, and15 plants are treated with nitrates, 16 with potash, 16 with phosphate, 7

93

94 5. The Principle of Inclusion and Exclusion

with nitrate and potash, 9 with nitrate and phosphate, 8 with potash andphosphate and 4 with all three. Now how many plants have been treated?If 32 plants were studied, how many received no treatment at all?

Problem 228.• Give a formula for the size of A ∪ B ∪ C in terms of the sizesof A, B, C and the intersections of these sets. (h)

5.1.2 Unions of an arbitrary number of sets

Problem 229.• Conjecture a formula for the size of a union of sets

A1 ∪ A2 ∪ · · · ∪ An =n⋃

i=1

Ai

in terms of the sizes of the sets Ai and their intersections.

The difficulty of generalizing Problem 228 to Problem 229 is not likely to be oneof being able to see what the right conjecture is, but of finding a good notation toexpress your conjecture. In fact, it would be easier for some people to express theconjecture in words than to express it in a notation. Here is some notation that willmake your task easier. Let us define ⋂

i:i∈I

Ai

to mean the intersection over all elements i in the set I of Ai . Thus⋂i:i∈{1,3,4,6}

= A1 ∩ A3 ∩ A4 ∩ A6. (5.1)

This kind of notation, consisting of an operator with a description underneathof the values of a dummy variable of interest to us, can be extended in many ways.For example ∑

I:I⊆{1,2,3,4}, |I |=2

| ∩i∈I Ai | = |A1 ∩ A2 | + |A1 ∩ A3 | + |A1 ∩ A4 |

+ |A2 ∩ A3 | + |A2 ∩ A4 | + |A3 ∩ A4 |. (5.2)

Problem 230.• Use notation something like that of Equation (5.1) and Equa-tion (5.2) to express the answer to Problem 229. Note there are many differ-ent correct ways to do this problem. Try to write down more than one andchoose the nicest one you can. Say why you chose it (because your view ofwhat makes a formula nice may be different from somebody else’s). The

5.1. The size of a union of sets 95

nicest formula won’t necessarily involve all the elements of Equations (5.1)and (5.2).

Problem 231.• A group of n students goes to a restaurant carrying back-packs. The manager invites everyone to check their backpack at the checkdesk and everyone does. While they are eating, a child playing in the checkroom randomly moves around the claim check stubs on the backpacks. Wewill try to compute the probability that, at the end of the meal, at least onestudent receives his or her own backpack. This probability is the fractionof the total number of ways to return the backpacks in which at least onestudent gets his or her own backpack back.

(a) What is the total number of ways to pass back the backpacks?

(b) In how many of the distributions of backpacks to students does at leastone student get his or her own backpack? (h)

(c) What is the probability that at least one student gets the correct back-pack?

(d) What is the probability that no student gets his or her own backpack?

(e)⇒ As the number of students becomes large, what does the probabilitythat no student gets the correct backpack approach?

Problem 231 is “classically” called the hatcheck problem; the name comesfrom substituting hats for backpacks. If is also sometimes called the derangementproblem. A derangement of an n-element set is a permutation of that set (thoughtof as a bĳection) that maps no element of the set to itself. One can think of a way ofhanding back the backpacks as a permutation f of the students: f (i) is the ownerof the backpack that student i receives. Then a derangement is a way to pass backthe backpacks so that no student gets his or her own.

5.1.3 The Principle of Inclusion and ExclusionThe formula you have given in Problem 230 is often called the principle of inclusionand exclusion for unions of sets. The reason is the pattern in which the formulafirst adds (includes) all the sizes of the sets, then subtracts (excludes) all the sizes ofthe intersections of two sets, then adds (includes) all the sizes of the intersectionsof three sets, and so on. Notice that we haven’t yet proved the principle. There area variety of proofs. Perhaps one of the most straightforward (though not the mostelegant) is an iductive proof that relies on the fact that

A1 ∪ A2 ∪ · · · ∪ An = (A1 ∪ A2 ∪ · · · ∪ An−1) ∪ An

and the formula for the size of a union of two sets.


Problem 232. Give a proof of your formula for the principle of inclusionand exclusion. (h)

Problem 233. We get a more elegant proof if we ask for a picture enumeratorfor A1 ∪ A2 ∪ · · · ∪ An . so let us assume A is a set with a picture function Pdefined on it and that each set Ai is a subset of A.

(a) By thinking about how we got the formula for the size of a union,write down instead a conjecture for the picture enumerator of a union.You could use notation like EP(

⋂i:i∈S Ai) for the picture enumerator

of the intersection of the sets Ai for i in a subset of S of [n].

(b) If x ∈ ⋃ni=1 Ai , what is the coefficient for P(x) in (the inclusion-

exclusion side of) your formula for EP(⋃n

i=1 Ai)? (h)

(c) If x !⋃n

i=1 Ai , what is the coefficient of P(x) in (the inclusion-exclusionside of) your formula for EP(

⋃ni=1 Ai)?

(d) How have you proved your conjecture for the picture enumerator ofthe union of the sets Ai?

(e) How can you get the formula for the principle of inclusion and exclu-sion from your formula for the picture enumerator of the union?

Problem 234. Frequently when we apply the principle of inclusion andexclusion, we will have a situation like that of part (d) of Problem 231.d.That is, we will have a set A and subsets A1 ,A2 , . . . ,An and we will wantthe size or the probability of the set of elements in A that are not in theunion. This set is known as the complement of the union of the Ais in A,and is denoted by A \⋃n

i=1 Ai , or if A is clear from context, by⋃n

i=1 Ai . Givethe fomula for

⋃ni=1 Ai . The principle of inclusion and exclusion generall

refers to both this formula and the one for the union.

We can find a very elegant way of writing the formula in Problem 234 if welet

⋂i:i∈∅ Ai = A. for this reason, if we have a family of subsets Ai of a set A, we

define1⋂

i:i∈∅ Ai = A.

1For those interested in logic and set theory, given a family of subsets Ai of a set A, we define⋂

i:i∈S Aito be the set of all members x of A that are in Ai for all i ∈ S. (Note that this allows x to be in some otherAjs as well.) Then if S = ∅, our intersection consists of all members x of A that satisfy the statement “ifi ∈ ∅, then x ∈ Ai .” But since the hypothesis of the “if-then” statement is false, the statement itself istrue for all x ∈ A. Therefor

⋂i:i∈∅ Ai = A.

5.2. Application of Inclusion and Exclusion 97

5.2 Application of Inclusion and Exclusion5.2.1 Multisets with restricted numbers of elements

Problem 235. In how many ways may we distribute k identical apples to nchildren so that no child gets more than four apples? Compare your resultwith your result in Problem 197 (h)

5.2.2 The Ménage Problem

Problem 236.⇒ A group of n married couples comes to a group discussionsession where they all sit around a round table. In how many ways can theysit so that no person is next to his or her spouse? (Note that two people ofthe same sex can sit next to each other.) (h)

Problem 237.⇒ ∗ A group of n married couples comes to a group discussionsession where they all sit around a round table. In how many ways can theysit so that no person is next to his or her spouse or a person of the same sex?This problem is called the ménage problem. (h)

5.2.3 Counting onto functions

Problem 238.• Given a function f from the k-element set K to the n-elementset [n], we say f is in the set Ai if f (x) " i for every x in K. How many ofthese sets does an onto function belong to? What is the number of functionsfrom a k-element set onto an n-element set?

Problem 239.⇒ Find a formula for the Stirling number (of the second kind)S(k , n). (h)

Problem 240. If we roll a die eight times, we get a sequence of 8 numbers,the number of dots on top on the first roll, the number on the second roll,and so on.

(a) What is the number of ways of rolling the die eight times so that eachof the numbers one through six appears at least once in our sequence?To get a numerical answer, you will likely need a computer algebrapackage.


(b) What is the probability that we get a sequence in which all six numbersbetween one and six appear? To get a numerical answer, you willlikely need a computer algebra package, programmable calculator, orspreadsheet.

(c) How many times do we have to roll the die to have probability atleast one half that all six numbers appear in our sequence. To an-swer this question, you will likely need a computer algebra package,programmable calculator, or spreadsheet.

5.2.4 The chromatic polynomial of a graphWe defined a graph to consist of set V of elements called vertices and a set E ofelements called edges such that each edge joins two vertices. A coloring of a graphby the elements of a set C (of colors) is an assignment of an element of C to eachvertex of the graph; that is, a function from the vertex set V of the graph to C.A coloring is called proper if for each edge joining two distinct vertices2, the twovertices it joins have different colors. You may have heard of the famous four colortheorem of graph theory that says if a graph may be drawn in the plane so that notwo edges cross (though they may touch at a vertex), then the graph has a propercoloring with four colors. Here we are interested in a different, though related,problem: namely, in how many ways may we properly color a graph (regardlessof whether it can be drawn in the plane or not) using k or fewer colors? Whenwe studied trees, we restricted ourselves to connected graphs. (Recall that a graphis connected if, for each pair of vertices, there is a walk between them.) Here,disconnected graphs will also be important to us. Given a graph which mightor might not be connected, we partition its vertices into blocks called connectedcomponents as follows. For each vertex v we put all vertices connected to it bya walk into a block together. Clearly each vertex is in at least one block, becausevertex v is connected to vertex v by the trivial walk consisting of the single vertexv and no edges. To have a partition, each vertex must be in one and only one block.To prove that we have defined a partition, suppose that vertex v is in the blocks B1

and B2. Then B1 is the set of all vertices connected by walks to some vertex v1 andB2 is the set of all vertices connected by walks to some vertex v2.

Problem 241.· (Relevant in Appendix C as well as this section.) Show thatB1 = B2.

Since B1 = B2, these two sets are the same block, and thus all blocks containingv are identical, so v is in only one block. Thus we have a partition of the vertexset, and the blocks of the partition are the connected components of the graph.Notice that the connected components depend on the edge set of the graph. Ifwe have a graph on the vertex set V with edge set E and another graph on the

2If a graph had a loop connecting a vertex to itself, that loop would connect a vertex to a vertex of thesame color. It is because of this that we only consider edges with two distinct vertices in our definition.

5.3. Deletion-Contraction and the Chromatic Polynomial 99

vertex set V with edge set E′, then these two graphs could have different connectedcomponents. It is traditional to use the Greek letter γ (gamma)3 to stand for thenumber of connected components of a graph; in particular, γ(V, E) stands for thenumber of connected components of the graph with vertex set V and edge set E.We are going to show how the principle of inclusion and exclusion may be used tocompute the number of ways to properly color a graph using colors from a set C ofc colors.

Problem 242.· Suppose we have a graph G with vertex set V and edge set E.Suppose F is a subset of E. Suppose we have a set C of c colors with whichto color the vertices.

(a) In terms of γ(V, F), in how many ways may we color the vertices of Gso that each edge in F connects two vertices of the same color? (h)

(b) Given a coloring of G, for each edge e in E, let us consider the propertythat the endpoints of e are colored the same color. Let us call thisproperty “property e.” In this way each set of properties can bethought of as a subset of E. What set of properties does a propercoloring have?

(c) Find a formula (which may involve summing over all subsets F ofthe edge set of the graph and using the number γ(V, F) of connectedcomponents of the graph with vertex set V and edge set F) for thenumber of proper colorings of G using colors in the set C. (h)

The formula you found in Problem 242.c is a formula that involves powers of c,and so it is a polynomial function of c. Thus it is called the chromatic polynomialof the graph G. Since we like to think about polynomials as having a variable xand we like to think of c as standing for some constant, people often use x as thenotation for the number of colors we are using to color G. Frequently people willuse χG(x) to stand for the number of way to color G with x colors, and call χG(x)the chromatic polynomial of G.

5.3 Deletion-Contraction and the Chromatic Polyno-mial

Problem 243.⇒ In Chapter 2 we introduced the deletion-contraction recur-rence for counting spanning trees of a graph. Figure out how the chromaticpolynomial of a graph is related to those resulting from deletion of an edgee and from contraction of that same edge e. Try to find a recurrence like theone for counting spanning trees that expresses the chromatic polynomialof a graph in terms of the chromatic polynomials of G − e and G/e for an

3The greek letter gamma is pronounced gam-uh, where gam rhymes with ham.


arbitrary edge e. Use this recurrence to give another proof that the numberof ways to color a graph with x colors is a polynomial function of x. (h)

Problem 244. Use the deletion-contraction recurrence to compute the chro-matic polynomial of the graph in Figure 5.3.1. (You can simplify yourcomputations by thinking about the effect on the chromatic polynomial ofdeleting an edge that is a loop, or deleting one of several edges between thesame two vertices.)

1 2

3

4

5

Figure 5.3.1: Agraph.

Problem 245.⇒

(a) In how many ways may you properly color the vertices of a path onn vertices with x colors? Describe any dependence of the chromaticpolynomial of a path on the number of vertices.

(b) (Not tremendously hard.) In how many ways may you properly colorthe vertices of a cycle on n vertices with x colors? Describe anydependence of the chromatic polynomial of a cycle on the number ofvertices.

Problem 246. In how many ways may you properly color the vertices of atree on n vertices with x colors? (h)

Problem 247.⇒ What do you observe about the signs of the coefficients of thechromatic polynomial of the graph in Figure 5.3.1? What about the signsof the coefficients of the chromatic polynomial of a path? Of a cycle? Of a


tree? Make a conjecture about the signs of the coefficients of a chromaticpolynomial and prove it.

5.4 Supplementary Problems1. Each person attending a party has been asked to bring a prize. The personplanning the party has arranged to give out exactly as many prizes as there areguests, but any person may win any number of prizes. If there are n guests, in howmany ways may the prizes be given out so that nobody gets the prize that he or shebrought?

2. There are m students attending a seminar in a room with n seats. The seminaris a long one, and in the middle the group takes a break. In how many ways maythe students return to the room and sit down so that nobody is in the same seat asbefore?

3. What is the number of ways to pass out k pieces of candy from an unlimitedsupply of identical candy to n children (where n is fixed) so that each child getsbetween three and six pieces of candy (inclusive)? If you have done Problem 1of Supplementary Problems 4.4, compare your answer in that problem with youranswer in this one.

4.⇒ In how many ways may k distinct books be arranged on n shelves so that noshelf gets more than m books?

5.⇒ Suppose that n children join hands in a circle for a game at nursery school. Thegame involves everyone falling down (and letting go). In how many ways may theyjoin hands in a circle again so that nobody is to the right of the same child that waspreviously to his or her right?

6.⇒ ∗ Suppose that n people link arms in a folk-dance and dance in a circle. Lateron they let go and dance some more, after which they link arms in a circle again.In how many ways can they link arms the second time so that no-one is next to aperson with whom he or she linked arms before.

7.⇒ ∗ (A challenge; the author has not tried to solve this one!) Redo Problem 6 in thecase that there are n men and n women and when people arrange themselves in acircle they do so alternating gender.

8.⇒ Suppose we take two graphs G1 and G2 with disjoint vertex sets, we choose onevertex on each graph, and connect these two graphs by an edge e to get a graphG12. How does the chromatic polynomial of G12 relate to those of G1 and G2?

Chapter 6

Groups acting on sets

6.1 Permutation GroupsUntil now we have thought of permutations mostly as ways of listing the elementsof a set. In this chapter we will find it very useful to think of permutations asfunctions. This will help us in using permutations to solve enumeration problemsthat cannot be solved by the quotient principle because they involve counting theblocks of a partition in which the blocks don’t have the same size. We begin bystudying the kinds of permutations that arise in situations where we have used thequotient principle in the past.

6.1.1 The rotations of a square

= identity

1

1�

2

2�

4

4

3

3

4

1�

1

2�

3

4

2

3

3

1�

4

2�

2

4

1

3

2

1�

3

2�

1

4

4

3

1

1�

2

2�

4

4

3

3

ρ�

ρ�

ρ� ρ� ρ�2 3� 4

0=

Figure 6.1.1: The four possible results of rotating a square and maintaining itsposition.

In Figure 6.1.1 we show a square with its four vertices labelled 1, 2, 3, and 4. Wehave also labeled the spot in the plane where each of these vertices falls with thesame label. Then we have shown the effect of rotating the square clockwise through90, 180, 270, and 360 degrees (which is the same as rotating through 0 degrees).Underneath each of the rotated squares we have named the function that carriesout the rotation. We use ρ, the Greek letter pronounced “row,” to stand for a 90degree clockwise rotation. We use ρ2 to stand for two 90 degree rotations, and so

103

104 6. Groups acting on sets

on. We can think of the function ρ as a function on the four element set1 {1, 2, 3, 4}.In particular, for any function ϕ (the Greek letter phi, usually pronounced “fee,”but sometimes “fie”) from the plane back to itself that may move the square aroundbut otherwise leaves it in the same place, we let ϕ(i) be the label of the place wherevertex previously in position i is now. Thus ρ(1) = 2, ρ(2) = 3, ρ(3) = 4 andρ(4) = 1. Notice that ρ is a permutation on the set {1, 2, 3, 4}.

Problem 248.• The composition f ◦ g of two functions f and g is defined byf ◦ g(x) = f (g(x)). Is ρ3 the composition of ρ and ρ2? Does the answerdepend on the order in which we write ρ and ρ2? How is ρ2 related to ρ?

Problem 249.• Is the composition of two permutations always a permuta-tion?

In Problem 248 you see that we can think of ρ2 ◦ ρ as the result of first rotatingby 90 degrees and then by another 180 degrees. In other words, the compositionof two rotations is the same thing as first doing one and then doing the other. Ofcourse there is nothing special about 90 degrees and 180 degrees. As long as wefirst do one rotation through a multiple of 90 degrees and then another rotationthrough a multiple of 90 degrees, the composition of these rotations is a rotationthrough a multiple of 90 degrees.

If we first rotate by 90 degrees and then by 270 degrees then we have rotatedby 360 degrees, which does nothing visible to the square. Thus we say that ρ4 isthe “identity function.” In general the identity function on a set S, denoted by ι(the Greek letter iota, pronounced eye-oh-ta) is the function that takes each elementof the set to itself. In symbols, ι(x) = x for every x in S. Of course the identityfunction on a set is a permutation of that set.

6.1.2 Groups of Permutations

Problem 250.• For any function ϕ from a set S to itself, we define ϕn (fornonnegative integers n) inductively by ϕ0 = ι and ϕn = ϕn−1 ◦ ϕ for everypositive integer n. If ϕ is a permutation, is ϕn a permutation? Based on yourexperience with previous inductive proofs, what do you expect ϕn ◦ ϕm tobe? What do you expect (ϕm)n to be? There is no need to prove these lasttwo answers are correct, for you have, in effect, already done so in Chapter 2.

Problem 251.• If we perform the composition ι ◦ ϕ for any function ϕ fromS to S, what function do we get? What if we perform the composition ϕ ◦ ι?

1What we are doing is restricting the rotation ρ to the set {1, 2, 3, 4}.

6.1. Permutation Groups 105

What you have observed about iota in Problem 251 is called the identity prop-erty of iota. In the context of permutations, people usually call the function ι “theidentity” rather than calling it “iota.”

Since rotating first by 90 degrees and then by 270 degrees has the same effectas doing nothing, we can think of the 270 degree rotation as undoing what the 90degree rotation does. For this reason we say that in the rotations of the square, ρ3 isthe “inverse” of ρ. In general, a function ϕ : T → S is called an inverse of a functionσ : S → T (the lower case Greek letter sigma) if ϕ ◦ σ = σ ◦ ϕ = ι. For a slowerintroduction to inverses and practice with them, see Section A.1.3 in Appendix A.Since a permutation is a bĳection, it has a unique inverse, as in Section A.1.3. Andsince the inverse of a bĳection is a bĳection (again, as in the Appendix), the inverseof a permutation is a permutation.

We use ϕ−1 to denote the inverse of the permutation ϕ. We’ve seen that therotations of the square are functions that return the square to its original positionbut may move the vertices to different places. In this way we create permutationsof the vertices of the square. We’ve observed three important properties of thesepermutations.

• (Identity Property) These permutations include the identity permutation.

• (Inverse Property) Whenever these permutations include ϕ, they also includeϕ−1.

• (Closure Property) Whenever these permutations include ϕ and σ, they alsoinclude ϕ ◦ σ.

A set of permutations with these three properties is called a permutation group2or a group of permutations. We call the group of permutations corresponding torotations of the square the rotation group of the square. There is a similar rotationgroup with n elements for any regular n-gon.

Problem 252.• If f : S → T, g : T → X, and h : X → Y, is h ◦ (g ◦ f ) =(h ◦ g) ◦ f ? What does this say about the status of the associative law

ρ ◦ (σ ◦ ϕ) = (ρ ◦ σ) ◦ ϕ

in a group of permutations?

Problem 253.•(a) How should we defineϕ−n for an elementϕ of a permutation group? (h)

(b) Will the two standard rules for exponents

am an = am+n and (am)n = amn

2The concept of a permutation group is a special case of the concept of a group that one studies inabstract algebra. When we refer to a group in what follows, if you know what groups are in the moreabstract sense, you may use the word in this way. If you do not know about groups in this more abstractsense, then you may assume we mean permutation group when we say group.


still hold if one or more of the exponents may be negative?

(c) What would we have to prove to show that the rules still hold?

(d) If the rules hold, give enough of the proof to show that you know howto do it; otherwise give a counterexample.

Problem 254.• If a finite set of permutations satisfies the closure property isit a permutation group? (h)

Problem 255.• There are three-dimensional geometric motions of the squarethat return it to its original position but move some of the vertices to otherpositions. For example, if we flip the square around a diagonal, most ofit moves out of the plane during the flip, but the square ends up in thesame place. Draw a figure like Figure 6.1.1 that shows all the possibleresults of such motions, including the ones shown in Figure 6.1.1. Do thecorresponding permutations form a group?

Problem 256. Let σ and ϕ be permutations.

(a) Why must σ ◦ ϕ have an inverse?

(b) Is (σ ◦ ϕ)−1 = σ−1ϕ−1? (Prove or give a counter-example.) (h)

(c) Is (σ ◦ ϕ)−1 = ϕ−1σ−1? (Prove or give a counter-example.)

Problem 257.• Explain why the set of all permutations of four elements is apermutation group. How many elements does this group have? This groupis called the symmetric group on four letters and is denoted by S4.

6.1.3 The symmetric groupIn general, the set of all permutations of an n-element set is a group. It is calledthe symmetric group on n letters. We don’t have nice geometric descriptions (likerotations) for all its elements, and it would be inconvenient to have to write downsomething like “Let σ(1) = 3, σ(2) = 1, σ(3) = 4, and σ(4) = 1” each time we needto introduce a new permutation. We introduce a new notation for permutationsthat allows us to denote them reasonably compactly and compose them reasonablyquickly. If σ is the permutation of {1, 2, 3, 4} given by σ(1) = 3, σ(2) = 1, σ(3) = 4


and σ(4) = 2, we write

σ =

(1 2 3 4

3 1 4 2

).

We call this notation the two row notation for permutations. In the two rownotation for a permutation of {a1 , a2 , . . . , an}, we write the numbers a1 through anin a one row and we write σ(a1) through σ(an) in a row right below, enclosing bothrows in parentheses. Notice that

(1 2 3 4

3 1 4 2

)=

(2 1 4 3

1 3 2 4

),

although the second ordering of the columns is rarely used.If ϕ is given by

ϕ =

(1 2 3 4

4 1 2 3

),

then, by applying the definition of composition of functions, we may compute σ◦ϕas shown in Figure 6.1.2.

1 2 3 4�3 1 4 2( ) ( )1 2 3 4�

4 1 2 3 ( )1 2 3 4�2 3 1 4

=

�

Figure 6.1.2: How to multiply permutations in two-row notation.

We don’t normally put the circle between two permutations in two row notationwhen we are composing them, and refer to the operation as multiplying the permu-tations, or as the product of the permutations. To see how Figure 6.1.2 illustratescomposition, notice that the arrow starting at 1 in ϕ goes to 4. Then from the 4 inϕ it goes to the 4 in σ and then to 2. This illustrates that ϕ(1) = 4 and σ(4) = 2, sothat σ(ϕ(1)) = 2.

Problem 258. For practice, compute(1 2 3 4 5

3 4 1 5 2

) (1 2 3 4 5

4 3 5 1 2

).

6.1.4 The dihedral groupWe found four permutations that correspond to rotations of the square. In Prob-lem 255 you found four permutations that correspond to flips of the square inspace. One flip fixes the vertices in the places labeled 1 and 3 and interchangesthe vertices in the places labeled 2 and 4. Let us denote it by ϕ1|3. One flip fixesthe vertices in the positions labeled 2 and 4 and interchanges those in the positionslabeled 1 and 3. Let us denote it by ϕ2|4. One flip interchanges the vertices in theplaces labeled 1 and 2 and also interchanges those in the places labeled 3 and 4.


Let us denote it by ϕ12|34. The fourth flip interchanges the vertices in the placeslabeled 1 and 4 and interchanges those in the places labeled 2 and 3. Let us denoteit by ϕ14|23. Notice that ϕ1|3 is a permutation that takes the vertex in place 1 to thevertex in place 1 and the vertex in place 3 to the vertex in place 3, while ϕ12|34 is apermutation that takes the edge between places 1 and 2 to the edge between places2 and 1 (which is the same edge) and takes the edge between places 3 and 4 to theedge between places 4 and 3 (which is the same edge). This should help to explainthe similarity in the notation for the two different kinds of flips.

Problem 259.• Write down the two-row notation for ρ3, ϕ2|4, ϕ12|34 andϕ2|4 ◦ ϕ12|34. Remember that σ(i) stands for the position where the vertexthat originated in position i is after we apply σ.

Problem 260. (You may have already done this problem in Problem 255, inwhich case you need not do it again!) In Problem 255, if a rigid motionof three-dimensional space returns the square to its original position, inhow many places can vertex number one land? Once the location of vertexnumber one is decided, how many possible locations are there for vertextwo? Once the locations of vertex one and vertex two are decided, howmany locations are there for vertex three? Answer the same question forvertex four. What does this say about the relationship between the fourrotations and four flips described above and the permutations you describedin Problem 255?

The four rotations and four flips of the square described before Problem 259form a group called the dihedral group of the square. Sometimes the group isdenoted D8 because it has eight elements, and sometimes the group is denoted byD4 because it deals with four vertices! Let us agree to use the notation D4 for thedihedral group of the square. There is a similar dihedral group, denoted by Dn , ofall the rigid motions of three-dimensional space that return a regular n-gon to itsoriginal position (but might put the vertices in different places.)

Problem 261. Another view of the dihedral group of the square is that itis the group of all distance preserving functions, also called isometries,from a square to itself. Notice that an isometry must be a bĳection. Anyrigid motion of the square preserves the distances between all points of thesquare. However, it is conceivable that there might be some isometries thatdo not arise from rigid motions. (We will see some later on in the case ofa cube.) Show that there are exactly eight isometires (distance preservingfunctions) from a square to itself. (h)


Problem 262. How many elements does the group Dn have? Prove that youare correct.

Problem 263. In Figure 6.1.3 we show a cube with the positions of its verticesand faces labeled. As with motions of the square, we let ϕ(x) be the labelof the place where vertex previously in position x is now.

Figure 6.1.3: A cube with the positions of its vertices and faces labelled. Thecurved arrows point to the positions that are blocked by the cube.

(a) Write in two row notation the permutation ρ of the vertices that corre-sponds to rotating the cube 90 degrees around a vertical axis throughthe faces t (for top) and u (for underneath). (Rotate in a right-handedfashion around this axis, meaning that vertex 6 goes to the back andvertex 8 comes to the front.)

(b) Write in two row notation the permutation ϕ that rotates the cube120 degrees around the diagonal from vertex 1 to vertex 7 and carriesvertex 8 to vertex 6.

(c) Compute the two row notation for ρ ◦ ϕ

(d) Is the permutation ρ ◦ ϕ a rotation of the cube around some axis? Ifso, say what the axis is and how many degrees we rotate around theaxis. If ρ ◦ ϕ is not a rotation, give a geometic description of it.

Problem 264.⇒ · How many permutations are in the group R? R is sometimescalled the “rotation group” of the cube. Can you justify this? (h)


Problem 265. As with a two-dimensional figure, it is possible to talk aboutisometries of a three-dimensional figure. These are distance preservingfunctions from the figure to itself. The function that reflects the cube inFigure 6.1.3 through a plane halfway between the bottom face and top faceexchanges the vertices 1 and 5, 2 and 6, 3 and 7, and 4 and 8 of the cube. Thisfunction preserves distances between points in the cube. However, it cannotbe achieved by a rigid motion of the cube because a rigid motion that takesvertex 1 to vertex 5, vertex 2 to vertex 6, vertex 3 to vertex 7, and vertex 4 tovertex 8 would not return the cube to its original location; rather it wouldput the bottom of the cube where its top previously was and would put therest of the cube above that square rather than below it.

(a) How many elements are there in the group of permutations of [8] thatcorrespond to isometries of the cube? (h)

(b) Is every permutation of [8] that corresponds to an isometry either arotation or a reflection? (h)

6.1.5 Group tables (Optional)We can always figure out the composition of two permutations of the same setby using the definition of composition, but if we are going to work with a givenpermutation group again and again, it is worth making the computations onceand recording them in a table. For example the group of rotations of the squaremay be represented as in Table 6.1.4. We list the elements of our group, with theidentity first, across the top of the table and down the left side of the table, usingthe same order both times. Then in the row labeled by the group element σ and thecolumn labelled by the group element ϕ, we write the composition σ◦ϕ, expressedin terms of the elements we have listed on the top and on the left side. Since agroup of permutations is closed under composition, the result σ ◦ ϕ will always beexpressible as one of these elements.

◦ ι ρ ρ2 ρ3

ι ι ρ ρ2 ρ3

ρ ρ ρ2 ρ3 ιρ2 ρ2 ρ3 ι ρρ3 ρ3 ι ρ ρ2

Table 6.1.4: The group table for the rotations of a square.

Problem 266. In Table 6.1.4, all the entries in a row (not including the firstentry, the one to the left of the line) are different. Will this be true in anygroup table for a permutation group? Why or why not? Also in Table 6.1.4all the entries in a column (not including the first entry, the one above the


line) are different. Will this be true in any group table for a permutationgroup? Why or why not?

Problem 267. In Table 6.1.4, every element of the group appears in everyrow (even if you don’t include the first element, the one before the line).Will this be true in any group table for a permutation group? Why or whynot? Also in Table 6.1.4 every element of the group appears in every column(even if you don’t include the first entry, the one before the line). Will thisbe true in any group table for a permutation group? Why or why not?

Problem 268.• Write down the group table for the dihedral group D4. Usethe ϕ notation described above to denote the flips. (Hints: Part of the tablehas already been written down. Will you need to think hard to write downthe last row? Will you need to think hard to write down the last column?When you multiply a product like ϕ1|3 ◦ ρ remember that we defined ϕ1|3to be the flip that fixes the vertex in position 1 and the vertex in position 3,not the one that fixes the vertex on the square labelled 1 and the vertex onthe square labelled 3.)

You may notice that the associative law, the identity property, and the inverseproperty are three of the most important rules that we use in regrouping paren-theses in algebraic expressions when solving equations. There is one property wehave not yet mentioned, the commutative law which would say that σ ◦ϕ = ϕ ◦ σ.It is easy to see from the group table of R4 that it satisfies the commutative law.

Problem 269. Does the commutative law hold in all permutation groups?

6.1.6 SubgroupsWe have seen that the dihedral group D4 contains a copy of the group of rotationsof the square. When one group G of permutations of a set S is a subset of anothergroup G′ of permutations of S, we say that G is a subgroup of G′.

Problem 270.• Find all subgroups of the group D4. (h)

Problem 271. Can you find subgroups of the symmetric group S4 with twoelements? Three elements? Four elements? Six elements? (For each positiveanswer, describe a subgroup. For each negative answer, explain why not.)


6.1.7 The cycle structure of a permutationThere is an even more efficient way to write down permutations. Notice that the

product in Figure 6.1.2 is(1 2 3 4

2 3 1 4

). We have drawn the directed graph of this

permutation in Figure 6.1.5.

1

234

Figure 6.1.5: The directed graph of a permutation .

You see that the graph has two directed cycles, the rather trivial one with vertex 4pointing to itself, and the nontrivial one with vertex 1 pointing to vertex 2 pointingto vertex 3 which points back to vertex 1. A permutation is called a cycle if its

digraph consists of exactly one cycle. Thus(1 2 3

2 3 1

)is a cycle but

(1 2 3 4

2 3 1 4

)

is not a cycle by our definition. We write (1 2 3) or (2 3 1) or (3 1 2) to stand for the

cycle σ =(1 2 3

2 3 1

).

We can describe cycles in another way as well. A cycle of the permutation σis a list (i σ(i) σ2(i) . . . σn(i)) that does not have repeated elements while the list(i σ(i) σ2(i) . . . σn(i)) σn+1(i)) does have repeated elements.

Problem 272. If the list (i σ(i) σ2(i) . . . σn(i)) does not have repeatedelements but the list (i σ(i) σ2(i) . . . σn(i) σn+1(i)) does have repeatedelements, then what is σn+1(i)? (h)

We say σ j(i) is an element of the cycle (i σ(i) σ2(i) . . . σn(i)). Notice that the casej = 0 means i is an element of the cycle. Notice also that if j > n, σ j(i) = σ j−n−1(i),so the distinct elements of the cycle are i, σ(i), σ2(i), through σn(i). We think ofthe cycle (i σ(i) σ2(i) . . . σn(i)) as representing the permutation σ restricted tothe set of elements of the cycle. We say that the cycles (i σ(i) σ2(i) . . . σn(i)) and( j σ( j) σ2( j) . . . σn( j)) are equivalent if there is an integer k such that j = σk(i).

Problem 273.• Find the cycles of the permutations ρ, ϕ1|3 and ϕ12|34 in thegroup D4.


Problem 274. Find the cycles of the permutation(1 2 3 4 5 6 7 8 9

3 4 6 2 9 7 1 5 8

).

Problem 275. If two cycles of σ have an element in common, what can wesay about them?

Problem 275 leads almost immediately to the following theorem.

Theorem 6.1.6. for each permutation σ of a set S, there is a unique partition of S each ofwhose blocks is the set of elements of a cycle of σ.

More informally, we may say that every permutation partitions its domain intodisjoint cycles. We call the set of cycles of a permutation the cycle decompositionof the permutation. Since the cycles of a permutation σ tell us σ(x) for every x inthe domain of σ, the cycle decomposition of a permutation completely determinesthe permutation. Using our informal language, we can express this idea in thefollowing corollary to Theorem 6.1.6.

Corollary 6.1.7. Every partition of a set S into cycles determins a unique permutation ofS.

Problem 276. Prove Theorem 6.1.6.

In Problems 273 and Problem 274 you found the cycle decomposition of typicalelements of the group D4 and of the permutation

(1 2 3 4 5 6 7 8 9

3 4 6 2 9 7 1 5 8

)

The group of all rotations of the square is simply the set of the four powersof the cycle ρ = (1 2 3 4). for this reason it is called a cyclic group3 and is oftendenoted by C4. Similarly, the rotation group of an n-gon is usually denoted Cn .

Problem 277.⇒ Write a recurrence for the number c(k , n) for the number ofpermutations of [k] that have exactly n cycles, including 1-cycles. Use itto write a table of c(k , n) for k between 1 and 7 inclusive. Can you find arelationship between c(k , n) and any of the other families of special numberssuch as binomial coefficients, Stirling numbers, Lah numbers, etc. we havestudied? If you find such a relationship, prove you are right. (h)

3The phrace cyclic group applies in a more general (but similar) situation. Namely the set of allpowers of any member of a group is called a cyclic group.


Problem 278.⇒ · (Relevant to Appendix C.) A permutation σ is called an in-volution if σ2 = ι. When you write an involution as a product of disjointcycles, what is special about the cycles?

6.2 Groups Acting on SetsWe defined the rotation group R4 and the dihedral group D4 as groups of permu-tations of the vertices of a square. These permutations represent rigid motions ofthe square in the plane and in three dimensional space respectively. The squarehas geometric features of interest other than its vertices; for example its diagonals,or its edges. Any geometric motion of the square that returns it to its originalposition takes each diagonal to a possibly different diagonal, and takes each edgeto a possibly different edge. In Figure 6.2.1 we show the results on the sides anddiagonals of the rotations of a square. The rotation group permutes the sides ofthe square and permutes the diagonals of the square as it rotates the square. Thus,we say the rotation group “acts” on the sides and diagonals of the square.

1 2

34

d13

d24

s4

s3

s2

s1 1 2

34

d24

d13

s3

s2

s1

s4

ρ

1 2

34

d13

d24

s2

s1

s4

s3

ρ2

1 2

34

d24

d13

s1

s4

s3

s2

ρ3

1 2

34

d13

d24

s4

s3

s2

s1

ρ4 = identity= ρ0

Figure 6.2.1: The results on the sides and diagonals of rotating the square

Problem 279.

(a) Write down the two-line notation for the permutation ρ that a 90degree rotation does to the sides of the square.

(b) Write down the two-line notation for the permutation ρ2 that a 180degree rotation does to the sides of the square.

(c) Is ρ2 = ρ ◦ ρ? Why or why not?

(d) Write down the two-line notation for the permutation ρ̂ that a 90degree rotation does to the diagonals d13 and d24 of the square.

(e) Write down the two-line notation for the permutation ρ̂2 that a 180degree rotation does to the diagonals d13 and d24 of the square.

(f) Is ρ̂2 = ρ̂ ◦ ρ̂? Why or why not? What familiar permutation is ρ̂2 inthis case?

6.2. Groups Acting on Sets 115

We have seen that the fact that we have defined a permutation group as thepermutations of some specific set doesn’t preclude us from thinking of the elementsof that group as permuting the elements of some other set as well. In order to keeptrack of which permutations of which set we are using to define our group andwhich other set is being permuted as well, we introduce some new language andnotation. We are going to say that the group D4 “acts” on the edges and diagonalsof a square and the group R of permutations of the vertices of a cube that arise fromrigid motions of the cube “acts” on the edges, faces, diagonals, etc. of the cube.

Problem 280.• In Figure 6.1.3 we show a cube with the positions of its verticesand faces labeled. As with motions of the square, we let we let ϕ(x) be thelabel of the place where vertex previously in position x is now.

(a) In Problem 263 we wrote in two row notation the permutation ρ ofthe vertices that corresponds to rotating the cube 90 degrees arounda vertical axis through the faces t (for top) and u (for underneath).(We rotated in a right-handed fashion around this axis, meaning thatvertex 6 goes to the back and vertex 8 comes to the front.) Write in tworow notation the permutation ρ of the faces that corresponds to thismember ρ of R.

(b) In Problem 263 we wrote in two row notation the permutation ϕ thatrotates the cube 120 degrees around the diagonal from vertex 1 tovertex 7 and carries vertex 8 to vertex 6. Write in two row notation theϕ of the faces that corresponds to this member of R.

(c) In Problem 263 we computed the two row notation for ρ ◦ ϕ. Nowcompute the two row notation for ρ ◦ ϕ (ρ was defined in Part 280.a),and write in two row notation the permutation ρ ◦ ϕ of the faces thatcorresponds to the action of the permutation ρ ◦ ϕ on the faces of thecube. (For this question it helps to think geometrically about whatmotion of the cube is carried out by ρ ◦ ϕ.) What do you observeabout ρ ◦ ϕ and ρ ◦ ϕ?

We say that a permutation group G acts on a set S if, for each member σ of Gthere is a permutation σ of S such that

σ ◦ ϕ = σ ◦ ϕ

for every member σ and ϕ of G. In Problem 280.c you saw one example of thiscondition. If we think intuitively of ρ and ϕ as motions in space, then followingthe action of ϕ by the action of ρ does give us the action of ρ ◦ ϕ. We can alsoreason directly with the permutations in the group R of rigid motions (rotations)of the cube to show that R acts on the faces of the cube.

Problem 281. Show that a group G of permutations of a set S acts on S withϕ = ϕ for all ϕ in G.


Problem 282.• The group D4 is a group of permutations of {1, 2, 3, 4} asin Problem 255. We are going to show in this problem how this groupacts on the two-element subsets of {1, 2, 3, 4}. In Problem 287 we will seea natural geometric interpretation of this action. In particular, for eachtwo-element subset {i , j} of {1, 2, 3, 4} and each member σ of D4 we defineσ({i , j}) = {σ(i), σ( j)}. Show that with this definition of σ, the group D4

acts on the two-element subsets of {1, 2, 3, 4}.

Problem 283.• Suppose that σ and ϕ are permutations in the group R ofrigid motions of the cube. We have argued already that each rigid motionsends a face to a face. Thus σ and ϕ both send the vertices on one face tothe vertices on another face. Let {h , i , j, k} be the set of labels next to thevertices on a face F.

(a) What are the vertices of the face F′ that F is sent to by ϕ?

(b) What are the vertices of the face F′′ that F′ is sent to by σ?

(c) What are the vertices of the face F′′′ that F is sent to by σ ◦ ϕ?

(d) How have you just shown that the group R acts on the faces?

6.2.1 Groups acting on colorings of setsRecall that when you were asked in Problem 45 to find the number of ways toplace two red beads and two blue beads at the corners of a square free to movein three-dimensional space, you were not able to apply the quotient principle toanswer the question. Instead you had to see that you could divide the set of sixlists of two Rs and two Bs into two sets, one of size two in which the Rs and Bsalternated and one of size four in which the two reds (and therefore the two blues)would be side-by-side on the square. Saying that the square is free to move in spaceis equivalent to saying that two arrangements of beads on the square are equivalentif a member of the dihedral group carries one arrangement to the other. Thus animportant ingredient in the analysis of such problems will be how a group canact on colorings of a set of vertices. We can describe the coloring of the square inFigure 6.2.2 as the function f with

f (1) = R, f (2) = R, f (3) = B, and f (4) = B,

but it is more compact and turns out to be more suggestive to represent the coloringin Figure 6.2.2 as the set of ordered pairs

(1, R), (2, R), (3, B), (4, B) (6.1)


4 3

1 2

B B

R R

Figure 6.2.2: The colored square with coloring {(1, R), (2, R), (3, B), (4, B)}

This gives us an explicity list of which colors are assigned to which vertex.4Then if we rotate the square through 90 degrees, we see that the set of orderedpairs becomes {

(ρ(1), R), (ρ(2), R), (ρ(3), B), (ρ(4), B)}

(6.2)which is the same as

{(2, R), (3, R), (4, B), (1, B)} .Or, in a more natural order,

{(1, B), (2, R), (3, R), (4, B)} . (6.3)

The reordering we did in (6.3) suggests yet another simplification of notation. Solong as we know we that the first elements of our pairs are labeled by the membersof [n] for some integer n and we are listing our pairs in increasing order by the firstcomponent, we can denote the coloring

{(1, B), (2, R), (3, R), (4, B)}

by BRRB. In the case where we have numbered the elements of the set S we arecoloring, we will call this list of colors of the elements of S in order the standardnotation for the coloring. We will call the ordering used in (6.3)the standardordering of the coloring.

Thus we have three natural ways to represent a coloring of a set: as a function,as a set of ordered pairs, and as a list. Different representations are useful fordifferent things. For example, the representation by ordered pairs will provide anatural way to define the action of a group on colorings of a set. Given a coloringas a function f , we denote the set of ordered pairs{

(x , f (x)) | x ∈ S}

,

suggestively as (S, f ) for short. We use f (1) f (2) · · · f (n) to stand for a particularcoloring (S, f ) in the standard notation.

Problem 284. Suppose now that instead of coloring the vertices of a square,we color its edges. We will use the shorthand 12, 23, 34, and 41 to stand forthe edges of the square between vertex 1 and vertex 2, vertex 2 and vertex

4The reader who has studied Appendix A will recognize that this set of ordered pairs is the relationof the function f , but we won’t need to make any specific references to the idea of a relation in whatfollows.


3, and so on. Then a coloring of the edges with 12 red, 23 blue, 34 red and41 blue can be represented as

{(12, R), (23, B), (34, R), (41, B)} . (6.4)

If ρ is the rotation through 90 degrees, then we have a permutation ρ actingon its edges. This permutation acts on the colorings to give us a permutationρ of the set of colorings.

(a) What is ρ of the coloring in (6.4)?

(b) What is ρ2 of the coloring in (6.4)?

If G is a group that acts on the set S, we define the action of G on the colorings(S, f ) by by

σ((S, f )) = σ({(x , f (x)) | x ∈ S

})=

{(σ(x), f (x)) | x ∈ S

}.. (6.5)

We have two bars over σ because σ is a permutation of one set that gives us apermutation σ of a second set, and then σ acts to give a permutation σ of a thid set,the set of colorings. For example, suppose we want to analyze colorings of the facesof a cube under the action of the rotation group of the cube as we have defined iton the vertices. Each vertex-permutation σ in the group gives a permutation σ ofthe faces of the cube. Then each permutation σ of the faces gives us a permutationσ of the colorings of the faces.

In the special case that G is a group of permutations of S rather than a groupacting on S, Equation (6.5) becomes

σ((S, f )) = σ({(x , f (x)) | x ∈ S}) = {(σ(x), f (x)) | x ∈ S}.

In the case where G is the rotation group of the square acting on the vertices of thesquare, the example of acting on a coloring by ρ that we saw in (6.3) is an exampleof this kind of action. In the standard notation, when we act on a coloring by σ, thecolor in position i moves to position σ(i).

Problem 285. Why does the action we have defined on colorings in Equa-tion (6.5) take a coloring to a coloring?

Problem 286. Show that if G is a group of permutations of a set S, and f isa coloring function on S, then the equation

σ({(x , f (x)) | x ∈ S}) = {(σ(x), f (x)) | x ∈ S}

defines an action of G on the colorings (S, f ) of S. (h)


6.2.2 Orbits

Problem 287.• Refer back to Problem 282 in answering the following ques-tions.

(a) What is the set of two element subsets that you get by computingσ({1, 2}) for all σ in D4?

(b) What is the multiset of two-element subsets that you get by computingσ({1, 2}) for all σin D4?

(c) What is the set of two-element subsets you get by computing σ({1, 3})for all σ in D4?

(d) What is the multiset of two-element subsets that you get by computingσ({1, 3}) for all σ in D4?

(e) Describe these two sets geometrically in terms of the square.

Problem 288.• This problem uses the notation for permutations in the dihe-dral group of the square introduced before Problem 259. What is the effectof a 180 degree rotation ρ2 on the diagonals of a square? What is the effectof the flip ϕ1|3 on the diagonals of a square? How many elements of D4

send each diagonal to itself? How many elements of D4 interchange thediagonals of a square?

In Problem 287 you saw that the action of the dihedral group D4 on two elementsubsets of {1, 2, 3, 4} seems to split them into two sets, one with two elementsand one with 4. We call these two sets the “orbits” of D4 acting on the twoelements subsets of {1, 2, 3, 4}. More generally, the orbit of a permutation group Gdetermined by an element x of a set S on which G acts is

{σ(x)|σ ∈ G},

and is denoted by Gx. In Problem 287 it was possible to have Gx = Gy. In fact inthat problem, Gx = Gy for every y in Gx.

Problem 289. Suppose a group acts on a set S. Could an element of S be intwo different orbits? (Say why or why not.) (h)

Problem 289 almost completes the proof of the following theorem.

Theorem 6.2.3. Suppose a group acts on a set S. The orbits of G form a partition of S.

It is probably worth pointing out that this theorem tells us that the orbit Gx isalso the orbit Gy for any element y of Gx.


Problem 290. Complete the proof of Theorem 6.2.3.

Notice that thinking in terms of orbits actually hides some information aboutthe action of our group. When we computed the multiset of all results of acting on{1, 2} with the elements of D4, we got an eight-element multiset containing eachside twice. When we computed the multiset of all results of acting on {1, 3} withthe elements of D4, we got an eight-element multiset containing each diagonal ofthe square four times. These multisets remind us that we are acting on our two-element sets with an eight-element group. The multiorbit of G determined by anelement x of S is the multiset

{σ(x) | σ ∈ G},

and is denoted by Gxmulti.When we used the quotient principle to count circular seating arrangements or

necklaces, we partitioned up a set of lists of people or beads into blocks of equivalentlists. In the case of seating n people around a round table, what made two listsequivalent was, in retrospect, the action of the rotation group Cn . In the case ofstringing n beads on a string to make a necklace, what made two lists equivalentwas the action of the dihedral group. Thus the blocks of our partitions were orbitsof the rotation group or the dihedral group, and we were counting the number oforbits of the group action. In Problem 45, we were not able to apply the quotientprinciple because we had blocks of different sizes. However, these blocks were stillorbits of the action of the group D4. And, even though the orbits have differentsizes, we expect that each orbit corresponds naturally to a multiorbit and that themultiorbits all have the same size. Thus if we had a version of the quotient rule fora union of multisets, we could hope to use it to count the number of multiorbits.

Problem 291.

(a) Find the orbit and multiorbit of D4 acting on the coloring

{(1, R), (2, R), (3, B), (4, B)},

or, in standard notation, RRBB of the vertices of a square.

(b) How many group elements map the coloring RRBB to itself? What isthe multiplicity of RRBB in its multiorbit?

(c) Find the orbit and multiorbit of D4 acting on the coloring

{(1, R), (2, B), (3, R), (4, B)}.

(d) How many elements of the group send the coloring RBRB to itself?What is the multiplicity of RBRB in its orbit?


Problem 292.

(a) If G is a group, how is the set {τσ | τ ∈ G} related to G?

(b) Use this to show that y is in the multiorbit Gxmulti if and only ifGxmulti = Gymulti.

Problem 292.b tells us that, when G acts on S, each element x of S is in one andonly one multiorbit. Since each orbit is a subset of a multiorbit and each element xin S is in one and only one orbit, this also tells us there is a bĳection between theorbits of G and the multiorbits of G, so that we have the same number of orbits asmultiorbits.

When a group acts on a set, a group element is said to fix an element of x ∈ S ifσ(x) = x. The set of all elements fixing an element x is denoted by (x).

Problem 293. Suppose a group G acts on a set S. What is special about thesubset (x) for an element x of S?

Problem 294.• Suppose a group G acts on a set S. What is the relationshipof the multiplicity of x ∈ S in its multiorbit and the size of (x)?

Problem 295. What can you say about relationships between the multiplic-ity of an element y in the multiorbit Gxmulti and the multiplicites of otherelements? Try to use this to get a relationship between the size of an orbitof G and the size of G. (h)

We suggested earlier that a quotient principle for multisets might prove useful.The quotient principle came from the sum principle, and we do not have a sumprinciple for multisets. Such a principle would say that the size of a union of disjointmultisets is the sum of their sizes. We have not yet defined the union of multisetsor disjoint multisets, because we haven’t needed the ideas until now. We define theunion of two multisets S and T to be the multiset in which the multiplicity of anelement x is the maximum5 of the multiplicity of x in S and its multiplicity in T .Similarly, the union of a family of multisets is defined by defining the multiplicity ofan element x to be the maximum of its multiplicities in the members of the family.Two multisets are said to be disjoint if no element is a member of both, that is, ifno element has multiplicity one or more in both. Since the size of a multiset is thesum of the multiplicities of its members, we immediately get the sum principle formultisets.

The size of a union of disjoint multisets is the sum of their sizes.

5We choose the maximum rather than the sum so that the union of sets is a special case of the unionof multisets.


Taking the multisets all to have the same size, we get the product principle formultisets.

The union of a set of m disjoint multisets, each of size n has size mn.

The quotient principle for multisets then follows immediately.

If a p-element multiset is a union of q disjoint multisets, each of size r,then q = p/r.

Problem 296.• How does the size of the union of the set of multiorbits of agroup G acting on a set S relate to the number of multiorbits and the size ofG?

Problem 297.• How does the size of the union of the set of multiorbits of agroup G acting on a set S relate to the numbers | (x)|?

Problem 298.• In Problems 296 and 297 you computed the size of the unionof the set of multiorbits of a group G acting on a set S in two different ways,getting two different expressions must be equal. Write the equation thatsays they are equal and solve for the number of multorbits, and thereforethe number of orbits.

6.2.3 The Cauchy-Frobenius-Burnside Theorem

Problem 299.• In Problem 298 you stated and proved a theorem that ex-presses the number of orbits in terms of the number of group elementsfixing each element of S. It is often easier to find the number of elementsfixed by a given group element than to find the number of group elementsfixing an element of S.

(a) For this purpose, how does the sum∑

x : x∈S | (x)| relate to thenumber of ordered pairs (σ, x) (with σ ∈ G and x ∈ S) such that σfixes x?

(b) Let χ(σ) denote the number of elements of S fixed by σ. How can thenumber of ordered pairs (σ, x) (with σ ∈ G and x ∈ S) such that σfixes x be computed from χ(G)? (It is ok to have a summation in youranswer.)

(c) What does this tell you about the number of orbits?


Problem 300. A second computation of the result of Problem 299 can bedone as follows.

(a) Let χ̂(σ, x) = 1 if σ(x) = x and let χ̂(σ, x) = 0 otherwise. Noticethat χ̂ is different from the χ in the previous problem, because it isa function of two variables. Use χ̂ to convert the single summationin Problem 298 into a double summation over elements x of S andelements σ of G.

(b) Reverse the order of the previous summation in order to convert it intoa single sum involving the function χ given by

χ(σ) = the number of elements of S left fixed by σ.

In Problem 299 you gave a formula for the number of orbits of a group G actingon a set X. This formula was first worked out by Cauchy in the case of the symmetricgroup, and then for more general groups by Frobenius. In his pioneering book onGroup Theory, Burnside used this result as a lemma, and while he attributed theresult to Cauchy and Frobenius in the first edition of his book, in later editions,he did not. Later on, other mathematicians who used his book named the result“Burnside’s Lemma," which is the name by which it is still most commonly known.Let us agree to call this result the Cauchy-Frobenius-Burnside Theorem, or CFBTheorem for short in a compromise between historical accuracy and common usage.

Problem 301.⇒ In how many ways may we string four (identical) red, six(identical) blue, and seven (identical) green beads on a necklace? (h)

Problem 302.⇒ If we have an unlimited supply of identical red beads andidentical blue beads, in how many ways may we string 17 of them on anecklace?

Problem 303.⇒ If we have five (identical) red, five (identical) blue, and five(identical) green beads, in how many ways may we string them on a neck-lace?

Problem 304.⇒ In how many ways may we paint the faces of a cube with sixdifferent colors, using all six?


Problem 305. In how many ways may we paint the faces of a cube with twocolors of paint? What if both colors must be used? (h)

Problem 306.⇒ In how many ways may we color the edges of a (regular)(2n + 1)-gon free to move around in the plane (so it cannot be flipped) if weuse red n times and blue n + 1 times? If this is a number you have seenbefore, identify it. (h)

Problem 307.⇒ ∗ In how many ways may we color the edges of a (regular)(2n + 1)-gon free to move in three-dimensional space so that n edges arecolored red and n + 1 edges are colored blue. Your answer may depend onwhether n is even or odd.

Problem 308.⇒ ∗ (Not unusually hard for someone who has worked on chro-matic polynomials.) How many different proper colorings with four colorsare there of the vertices of a graph which is cycle on five vertices? (If we getone coloring by rotating or flipping another one, they aren’t really different.)

Problem 309.⇒ ∗ How many different proper colorings with four colors arethere of the graph in Figure 6.2.4? Two graphs are the same if we canredraw one of the graphs, not changing the vertex set or edge set, so that itis identical to the other one. This is equivalent to permuting the vertices insome way so that when we apply the permutation to the endpoints of theedges to get a new edge set, the new edge set is equal to the old one. Sucha permutation is called an automorphism of the graph. Thus two coloringsare different if there is no automorphism of the graph that carries one to theother one.

1� 2�

3

4�5�

6

Figure 6.2.4: A graph on six vertices.

(h)

6.3. Pólya-Redfield Enumeration Theory 125

6.3 Pólya-Redfield Enumeration TheoryGeorge Pólya and Robert Redfield independently developed a theory of generatingfunctions that describe the action of a group G on functions from a set S to a set Twhen we know the action of G on S. Pólya’s work on the subject is very accessiblein its exposition, and so the subject has become popularly known as Pólya theory,though Pólya-Redfield theory would be a better name. In this section we developthe elements of this theory.

The idea of coloring a set S has many applications. For example, the set S mightbe the positions in a hydrocarbon molecule which are occupied by hydrogen, andthe group could be the group of spatial symmetries of the molecule (that is, thegroup of permutations of the atoms of the molecule that move the molecule aroundso that in its final position the molecule cannot be distinguished from the originalmolecule). The colors could then be radicals (including hydrogen itself) that wecould substitute for each hydrogen position in the molecule. Then the numberof orbits of colorings is the number of chemically different compounds we couldcreate by using these substitutions.6

In Figure 6.3.1 we show two different ways to substitute the OH radical for ahydrogen atom in the chemical diagram we gave for butane in Chapter 2. We havecolored one vertex of degree 1 with the radical OH and the rest with the atom H.There are only two distinct ways to do this, as the OH must either connect to an“end” C or a “middle” C. This shows that there are two different forms, calledisomers of the compound shown. This compound is known as butyl alcohol.

Figure 6.3.1: The two different isomers of butyl alcohol.

So think intuitively about some “figure” that has places to be colored. (Thinkof the faces of a cube, the beads on a necklace, circles at the vertices of an n-gon,etc.) How can we picture the coloring? If we number the places to be colored,say 1 to n, then a function from [n] to the colors is exactly our coloring; if ourcolors are blue, green and red, then BBGRRGBG describes a typical coloring of 8such places. Unless the places are somehow “naturally” numbered, this idea of acoloring imposes structure that is not really there. Even if the structure is there,visualizing our colorings in this way doesn’t “pull together” any common features

6There is a fascinating subtle issue of what makes two molecules different. For example, supposewe have a molecule in the form of a cube, with one atom at each vertex. If we interchange the top andbottom faces of the cube, each atom is still connected to exactly the same atoms as before. However wecannot achieve this permutation of the vertices by a member of the rotation group of the cube. It couldwell be that the two versions of the molecule interact with other molecules in different ways, in whichcase we would consider them chemically different. On the other hand if the two versions interact withother molecules in the same way, we would have no reason to consider them chemically different. Thiskind of symmetry is an example of what is called chirality in chemistry.


of different colorings; we are simply visualizing all possible functions. We havea group (think of it as symmetries of the figure you are imagining) that acts onthe places. That group then acts in a natural way on the colorings of the placesand we are interested in orbits of the colorings. Thus we want a picture that pullstogether the common features of the colorings in an orbit. One way to pull togethersimilarities of colorings would be to let the letters we are using as pictures of colorscommute as we did with our pictures in Chapter 4; then our picture BBGRRGBGbecomes B3G3R2, so our picture now records simply how many times we use eachcolor. If you think about how we defined the action of a group on a set of functions,you will see that a group element won’t change how many times each color is used;it simply moves colors to different places. Thus the picture we now have of a givencoloring is an equally appropriate picture for each coloring in an orbit. One naturalquestion for us to ask is “How many orbits have a given picture?”

Problem 310. Suppose we draw identical circles at the vertices of a regularhexagon. Suppose we color these circles with two colors, red and blue.

(a) In how many ways may we color the set {1, 2, 3, 4, 5, 6} using the colorsred and blue?

(b) These colorings are partitioned into orbits by the action of the rotationgroup on the hexagon. Using our standard notation, write down allthese orbits and observe how many orbits have each picture, assum-ing the picture of a coloring is the product of commuting variablesrepresenting the colors.

(c) Using the picture function of the previous part, write down the pictureenumerator for the orbits of colorings of the vertices of a hexagonunder the action of the rotation group.

In Problem c we saw a picture enumerator for pictures of orbits of the action ofa group on colorings. As above, we can ask how many orbits of the colorings haveany given picture. We can think of a multivariable generating function in whichthe letters we use to picture individual colors are the variables, and the coefficientof a picture is the number of orbits with that picture. Such a generating function isan answer to our natural question, and so it is this sort of generating function wewill seek. Since the CFB theorem was our primary tool for saying how many orbitswe have, it makes sense to think about whether the CFB theorem has an analog interms of pictures of orbits.

6.3.1 The Orbit-Fixed Point Theorem

Problem 311.• Suppose now we have a group G acting on a set and we havea picture function on that set with the additional feature that for each orbitof the group, all its elements have the same picture. In this circumstancewe define the picture of an orbit or multiorbit to be the picture of any oneof its members. The orbit enumerator (G, S) is the sum of the pictures


of the orbits. (Note that this is the same as the sum of the pictures ofthe multiorbits.) The fixed point enumerator (G, S) is the sum of thepictures of each of the fixed points of each of the elements of G. We aregoing to construct a generating function analog of the CFB theorem. Themain idea of the proof of the CFB theorem was to try to compute in twodifferent ways the number of elements (i.e. the sum of all the multiplicitiesof the elements) in the union of all the multiorbits of a group acting on aset. Suppose instead we try to compute the sum of all the pictures of allthe elements in the union of the multiorbits of a group acting on a set. Bythinking about how this sum relates to (G, S) and (G, S), find ananalog of the CFB theorem that relates these two enumerators. State andprove this theorem.

We call the theorem of Problem 311 the Orbit-Fixed Point Theorem. In or-der to apply the Orbit-Fixed Point Theorem, we need some facts about pictureenumerators.

Problem 312.• Suppose that P1 and P2 are picture functions on sets S1 and S2

in the sense of Section 4.1.2. Define P on S1×S2 by P(x1 , x2) = P1(x1)P2(x2).How are EP1 , EP1 , and EP related? (You may have already done this problemin another context!)

Problem 313.• Suppose Pi is a picture function on a set Si for i = 1, . . . , k.We define the picture of a k-tuple (x1 , x2 , . . . , xk) to be the product of thepictures of its elements, i.e.

P̂((x1 , x2 , . . . , xk)) =k∏

i=1

Pi(xi).

How does the picture enumerator EP̂ of the set S1×S2× · · ·×Sk of all k-tupleswith xi ∈ S relate to the picture enumerators of the sets Si? In the specialcase that Si = S for all i and Pi = P for all i, what is EP̂(S

k)?

Problem 314.• Use the Orbit-Fixed Point Theorem to determine the OrbitEnumerator for the colorings, with two colors (red and blue), of six circlesplaced at the vertices of a hexagon which is free to move in the plane.Compare the coefficients of the resulting polynomial with the various orbitsyou found in Problem 310.


Problem 315. Find the generating function (in variables R, B) for coloringsof the faces of a cube with two colors (red and blue). What does the gen-erating function tell you about the number of ways to color the cube (up tospatial movement) with various combinations of the two colors.

6.3.2 The Pólya-Redfield TheoremPólya’s (and Redfield’s) famed enumeration theorem deals with situations such asthose in Problems 314 and Problem 315 in which we want a generating function forthe set of all functions from a set S to a set T on which a picture function is defined,and the picture of a function is the product of the pictures of its multiset of values.The point of the next series of problems is to analyze the solution to Problems 314and Problem 315 in order to see what Pólya and Redfield saw (though they didn’tsee it in this notation or using this terminology).

Problem 316.• In Problem 314 we have four kinds of group elements: theidentity (which fixes every coloring), the rotations through 60 or 300 degrees,the rotations through 120 and 240 degrees, and the rotation through 180degrees. The fixed point enumerator for the rotation group acting on thefunctions is by definition the sum of the fixed point enumerators of coloringsfixed by the identity, of colorings fixed by 60 or 300 degree rotations, ofcolorings fixed by 120 or 240 degree rotations, and of colorings fixed by the180 degree rotation. Write down each of these enumerators (one for eachkind of permutation) individually and factor each one (over the integers) ascompletely as you can.

Problem 317.• In Problem 315 we have five different kinds of group ele-ments, and the fixed point enumerator is the sum of the fixed point enumer-ators of each of these kinds of group elements. For each kind of element,write down the fixed point enumerator for the elements of that kind. Factorthe enumerators as completely as you can.

Problem 318.• In Problem 316, each “kind” of group element has a “kind” ofcycle structure. For example, a rotation through 180 degrees has three cyclesof size two. What kind of cycle structure does a rotation through 60 or 300degrees have? What kind of cycle structure does a rotation through 120 or240 degrees have? Discuss the relationship between the cycle structures andthe factored enumerators of fixed points of the permutations in Problem 316.


Recall that we said that a group of permutations acts on a set if, for each memberσ of G there is a bĳection σ of S such that

σ ◦ ϕ = σ ◦ ϕfor every member σ and ϕ of G. Since σ is a bĳection of S to itself, it is in fact apermutation of S. Thus σ has a cycle structure (that is, it is a product of disjointcycles) as a permutation of S (in addition to whatever its cycle structure is in theoriginal permutation group G).

Problem 319.• In Problem 317, each “kind” of group element has a “kind”of cycle structure in the action of the rotation group of the cube on the facesof the cube. For example, a rotation of the cube through 180 degrees arounda vertical axis through the centers of the top and bottom faces has two cyclesof size two and two cycles of size one. How many such rotations does thegroup have? What are the other “kinds” of group elements, and what aretheir cycle structures? Discuss the relationship between the cycle structureand the factored enumerator in Problem 317.

Problem 320.• The usual way of describing the Pólya-Redfield enumerationtheorem involves the “cycle indicator” or “cycle index” of a group acting ona set. Suppose we have a group G acting on a finite set S. Since each groupelement σ gives us a permutation σ of S, as such it has a decomposition intodisjoint cycles as a permutation of S. Suppose σ has c1 cycles of size 1, c2cycles of size 2, ..., cn cycles of size n. Then the cycle monomial of σ is

z(σ) = zc11 zc2

2 · · · zcnn .

The cycle indicator or cycle index of G acting on S is

Z(G, S) =1

|G |∑σ:σ∈G

z(σ).

(a) What is the cycle index for the group D6 acting on the vertices of ahexagon?

(b) What is the cycle index for the group of rotations of the cube actingon the faces of the cube?

Problem 321. How can you compute the Orbit Enumerator of G acting onfunctions from S to a finite set T from the cycle index of G acting on S? (Use t,thought of as a variable, as the picture of an element t of T.) State and provethe relevant theorem! This is Pólya’s and Redfield’s famous enumerationtheorem.


Problem 322.⇒ Suppose we make a necklace by stringing 12 pieces of brightlycolored plastic tubing onto a string and fastening the ends of the stringtogether. We have ample supplies blue, green, red, and yellow tubingavailable. Give a generating function in which the coefficient of BiGjRkYh

is the number of necklaces we can make with i blues, j greens, k reds, andh yellows. How many terms would this generating function have if youexpanded it in terms of powers of B, G, R, and Y? Does it make sense todo this expansion? How many of these necklaces have 3 blues, 3 greens, 2reds, and 4 yellows?

Problem 323. What should we substitute for the variables representing col-orings in the orbit enumerator of G acting on the set of colorings of S by a setT of colors in order to compute the total number of orbits of G acting on theset of colorings? What should we substitute into the variables in the cycleindex of a group G acting on a set S in order to compute the total number oforbits of G acting on the colorings of S by a set T? Find the number of waysto color the faces of a cube with four colors.

Problem 324.⇒ We have red, green, and blue sticks all of the same length,with a dozen sticks of each color. We are going to make the skeleton of acube by taking eight identical lumps of modeling clay and pushing threesticks into each lump so that the lumps become the vertices of the cube.(Clearly we won’t need all the sticks!) In how many different ways couldwe make our cube? How many cubes have four edges of each color? Howmany have two red, four green, and six blue edges?

Problem 325.⇒ How many cubes can we make in Problem 324 if the lumpsof modelling clay can be any of four colors?

1� 2�

3

4�5�

6

Figure 6.3.2: A possible computer network.


Problem 326.⇒ In Figure 6.3.2 we see a graph with six vertices. Supposewe have three different kinds of computers that can be placed at the sixvertices of the graph to form a network. In how many different ways maythe computers be placed? (Two graphs are not different if we can redrawone of the graphs so that it is identical to the other one.) This is equivalent topermuting the vertices in some way so that when we apply the permutationto the endpoints of the edges to get a new edge set, the new edge set is equalto the old one. Such a permutation is called an automorphism of the graph.Then two computer placements are the same if there is an automorphismof the graph that carries one to the other. (h)

Problem 327.⇒ Two simple graphs on the set [n] = {1, 2, . . . , n} with edgesets E and E′ (which we think of a sets of two-element sets for this problem)are said to be isomorphic if there is a permutation σ of [n] which, in itsaction of two-element sets, carries E to E′. We say two graphs are different ifthey are not isomorphic. Thus the number of different graphs is the numberof orbits of the set of all two-element subsets of [n] under the action of thegroup Sn . We can represent an edge set by its characteristic function (as inproblem 33). That is we define

χE({u , v}) ={1 if {u , v} ∈ E0 otherwise.

Thus we can think of the set of graphs as a set of colorings with colors 0 and1 of the set of all two-element subsets of [n]. The number of different graphswith vertex set [n] is thus the number of orbits of this set of colorings underthe action of the symmetric group Sn on the set of two-element subsets of[n]. Use this to find the number of different graphs on five vertices. (h)

6.4 Supplementary Problems1. Show that a function from S to T has an inverse (defined on T) if and only if itis a bĳection.

2. How many elements are in the dihedral group D3? The symmetric group S3?What can you conclude about D3 and S3?

3. A tetrahedron is a thee dimensional geometric figure with four vertices, sixedges, and four triangular faces. Suppose we start with a tetrahedron in spaceand consider the set of all permutations of the vertices of the tetrahedron thatcorrespond to moving the tetrahedron in space and returning it to its originallocation, perhaps with the vertices in different places.(a) Explain why these permutations form a group.(b) What is the size of this group?


(c) Write down in two-row notation a permutation that is not in this group.

4. Find a three-element subgroup of the group S3. Can you find a different three-element subgroup of S3?

5. Prove true or demonstrate false with a counterexample: “In a permutationgroup, (σϕ)n = σnϕn .”

6. If a group G acts on a set S, and if σ(x) = y, is there anything interesting we cansay about the subgroups (x) and (y)?

7.(a) If a group G acts on a set S, does σ( f ) = f ◦ σ define a group action on the

functions from S to a set T? Why or why not?(b) If a group G acts on a set S, does σ( f ) = f ◦ σ−1 define a group action on the

functions from S to a set T ? Why or why not?(c) Is either of the possible group actions essentially the same as the action we

described on colorings of a set, or is that an entirely different action?

8. Find the number of ways to color the faces of a tetrahedron with two colors.

9. Find the number of ways to color the faces of a tetrahedron with four colors sothat each color is used.

10. Find the cycle index of the group of spatial symmetries of the tetrahedronacting on the vertices. Find the cycle index for the same group acting on the faces.

11. Find the generating function for the number of ways to color the faces of thetetrahedron with red, blue, green and yellow.

12.⇒ Find the generating function for the number of ways to color the faces of a cubewith four colors so that all four colors are used.

13.⇒ How many different graphs are there on six vertices with seven edges?

14.⇒ Show that if H is a subgroup of the group G, then H acts on G by σ(τ) = σ ◦ τfor all σ in H and τ in G. What is the size of an orbit of this action? How does thesize of a subgroup of a group relate to the size of the group?

Appendix A

Relations

A.1 Relations as sets of Ordered PairsA.1.1 The relation of a function

Problem 328. Consider the functions from S = {−2,−1, 0, 1, 2} to T ={1, 2, 3, 4, 5} defined by f (x) = x + 3, and g(x) = x5 − 5x3 + 5x + 3. Writedown the set of ordered pairs (x , f (x)) for x ∈ S and the set of ordered pairs(x , g(x)) for x ∈ S. Are the two functions the same or different?

Problem 328 points out how two functions which appear to be different areactually the same on some domain of interest to us. Most of the time when we arethinking about functions it is fine to think of a function casually as a relationshipbetween two sets. In Problem 328 the set of ordered pairs you wrote down foreach function is called the relation of the function. When we want to distinguishbetween the casual and the careful in talking about relationships, our casual termwill be “relationship” and our careful term will be “relation.” So relation is atechnical word in mathematics, and as such it has a technical definition. A relationfrom a set S to a set T is a set of ordered pairs whose first elements are in S andwhose second elements are in T. Another way to say this is that a relation from Sto T is a subset of S × T.

A typical way to define a function f from a set S, called the domain of thefunction, to a set T, called the range, is that f is a relationship between S to T thatrelates one and only one member of T to each element of X. We use f (x) to standfor the element of T that is related to the element x of S. If we wanted to makeour definition more precise, we could substitute the word “relation” for the word“relationship” and we would have a more precise definition. For our purposes,you can choose whichever definition you prefer. However, in any case, there is arelation associated with each function. As we said above, the relation of a functionf : S → T (which is the standard shorthand for “ f is a function from S to T” andis usually read as f maps S to T) is the set of all ordered pairs (x , f (x)) such that xis in S.

133

134 A. Relations

Problem 329. Here are some questions that will help you get used to theformal idea of a relation and the related formal idea of a function. S willstand for a set of size s and T will stand for a set of size t.

(a) What is the size of the largest relation from S to T?

(b) What is the size of the smallest relation from S to T?

(c) The relation of a function f : S → T is the set of all ordered pairs(x , f (x)) with x ∈ S. What is the size of the relation of a functionfrom S to T? That is, how many ordered pairs are in the relation of afunction from S to T? (h)

(d) We say f is a one-to-one function or injection from S to T if eachmember of S is related to a different element of T. How many differentelements must appear as second elements of the ordered pairs in therelation of a one-to-one function (injection) from S to T?

(e) A function f : S → T is called an onto function or surjection if eachelement of T is f (x) for some x ∈ S What is the minimum size that Scan have if there is a surjection from S to T?

Problem 330. When f is a function from S to T, the sets S and T play a bigrole in determining whether a function is one-to-one or onto (as defined inProblem 329). For the remainder of this problem, let S and T stand for theset of nonnegative real numbers.

(a) If f : S → T is given by f (x) = x2, is f one-to-one? Is f onto?

(b) Now assume S′ is the set of all real numbers and g : S′ → T is givenby g(x) = x2. Is g one-to-one? Is g onto?

(c) Assume that T′ is the set of all real numbers and h : S → T′ is givenby h(x) = x2. Is h one-to-one? Is h onto?

(d) And if the function j : S′ → T′ is given by j(x) = x2, is j one-to-one?Is j onto?

Problem 331. If f : S → T is a function, we say that f maps x to y as anotherway to say that f (x) = y. Suppose S = T = {1, 2, 3}. Give a function fromS to T that is not onto. Notice that two different members of S have mappedto the same element of T. Thus when we say that f associates one and onlyone element of T to each element of S, it is quite possible that the one andonly one element f (1) that f maps 1 to is exactly the same as the one andonly one element f (2) that f maps 2 to.

A.1. Relations as sets of Ordered Pairs 135

A.1.2 Directed graphsWe visualize numerical functions like f (x) = x2 with their graphs in Cartesian co-ordinate systems. We will call these kinds of graphs coordinate graphs to distinguishthem from other kinds of graphs used to visualize relations that are non-numerical.

a�

b�

c�

d�

Figure A.1.1: The alphabet digraph.

In Figure A.1.1 we illustrate another kind of graph, a “directed graph” or “di-graph” of the “comes before in alphabetical order" relation on the letters a, b, c,and d. To draw a directed graph of a relation on a set S, we draw a circle (or dot,if we prefer), which we call a vertex, for each element of the set, we usually labelthe vertex with the set element it corresponds to, and we draw an arrow from thevertex for a to that for b if a is related to b, that is, if the ordered pair (a , b) is in ourrelation. We call such an arrow an edge or a directed edge. We draw the arrowfrom a to b, for example, because a comes before b in alphabetical order. We try tochoose the locations where we draw our vertices so that the arrows capture whatwe are trying to illustrate as well as possible. Sometimes this entails redrawingour directed graph several times until we think the arrows capture the relationshipwell.

We also draw digraphs for relations from a set S to a set T; we simply drawvertices for the elements of S (usually in a row) and vertices for the elements of T(usually in a parallel row) draw an arrow from x in S to y in T if x is related to y.Notice that instead of referring to the vertex representing x, we simply referred tox. This is a common shorthand. Here are some exercises just to practice drawingdigraphs.

Problem 332. Draw the digraph of the “is a proper subset of” relation onthe set of subsets of a two element set. How many arrows would you havehad to draw if this problem asked you to draw the digraph for the subsetsof a three-element set? (h)

We also draw digraphs for relations from finite set S to a finite set T; we simplydraw vertices for the elements of S (usually in a row) and vertices for the elementsof T (usually in a parallel row) and draw an arrow from x in S to y in T if x isrelated to y. Notice that instead of referring to the vertex representing x, we simplyreferred to x. This is a common shorthand.

136 A. Relations

Problem 333. Draw the digraph of the relation from the set {A, M, P, S} tothe set {Sam, Mary, Pat, Ann, Polly, Sarah} given by “is the first letter of.”

Problem 334. Draw the digraph of the relation from the set {Sam, Mary,Pat, Ann, Polly, Sarah} to the set {A, M, P, S} given by “has as its first letter.”

Problem 335. Draw the digraph of the relation on the set {Sam, Mary, Pat,Ann, Polly, Sarah} given by “has the same first letter as.”

A.1.3 Digraphs of Functions

Problem 336. When we draw the digraph of a function f , we draw an arrowfrom the vertex representing x to the vertex representing f (x). One of therelations you considered in Problems 333 and Problem 334 is the relation ofa function.

(a) Which relation is the relation of a function?

(b) How does the digraph help you visualize that one relation is a functionand the other is not?

Problem 337. Digraphs of functions help us to visualize whether or notthey are onto or one-to-one. For example, let both S and T be the set{−2,−1, 0, 1, 2} and let S′ and T′ be the set {0, 1, 2}. Let f (x) = 2 − |x |.

(a) Draw the digraph of the function f assuming its domain is S and itsrange is T. Use the digraph to explain why or why not this functionmaps S onto T.

(b) Use the digraph of the previous part to explain whether or not thefunction is one-to one.

(c) Draw the digraph of the function f assuming its domain is S and itsrange is T′. Use the digraph to explain whether or not the function isonto.

(d) Use the digraph of the previous part to explain whether or not thefunction is one-to-one.

(e) Draw the digraph of the function f assuming its domain is S′ and itsrange is T′. Use the digraph to explain whether the function is onto.

A.1. Relations as sets of Ordered Pairs 137

(f) Use the digraph of the previous part to explain whether the functionis one-to-one.

(g) Suppose the function f has domain S′ and range T. Draw the digraphof f and use it to explain whether f is onto.

(h) Use the digraph of the previous part to explain whether f is one-to-one.

A one-to-one function from a set X onto a set Y is frequently called a bĳection,especially in combinatorics. Your work in Problem 337 should show you that adigraph is the digraph of a bĳection from X to Y

• if the vertices of the digraph represent the elements of X and Y,

• if each vertex representing an element of X has one and only one arrowleaving it, and

• each vertex representing an element of Y has one and only one arrow enteringit.

Problem 338. If we reverse all the arrows in the digraph of a bĳection f , weget the digraph of another function g. Is g a bĳection? What is f (g(x))?What is g( f (x))?

If f is a function from S to T, if g is a function from T to S, and if f (g(x)) = x foreach x in T and g( f (x)) = x for each x in S, then we say that g is an inverse of f(and f is an inverse of g).

More generally, if f is a function from a set R to a set S, and g is a function fromS to T, then we define a new function f ◦ g, called the composition of f and g , byf ◦ g(x) = f (g(x)). Composition of functions is a particularly important operatioin subjects such as calculus, where we represent a function like h(x) =

√x2 + 1 as

the composition of the square root function and the square and add one functionin order to use the chain rule to take the derivative of h.

The function ι (the Greek letter iota is pronounced eye-oh-ta) from a set S toitself, given by the rule ι(x) = x for every x in S, is called the identity function onS. If f is a function from S to T and g is a function from T to S such that g( f (x)) = xfor every x in S, we can express this by saying that g ◦ f = ι, where ι is the identityfunction of S. Saying that f (g(x)) = x is the same as saying that f ◦ g = ι, whereι stands for the identity function on T. We use the same letter for the identityfunction on two different sets when we can use context to tell us on which set theidentity function is being defined.

Problem 339. If f is a function from S to T and g is a function from T to Ssuch that g( f (x)) = x, how can we tell from context that g ◦ f is the identityfunction on S and not the identity function on T? (h)

138 A. Relations

Problem 340. Explain why a function that has an inverse must be a bĳection.

Problem 341. Is it true that the inverse of a bĳection is a bĳection?

Problem 342. If g and h are inverse of f , then what can we say about g andh?

Problem 343. Explain why a bĳection must have an inverse.

Since a function with an inverse has exactly one inverse g, we call g the inverseof f . From now on, when f has an inverse, we shall denote its inverse by f −1. Thusf ( f −1(x)) = x and f −1( f (x)) = x. Equivalenetly f ◦ f −1 = ι and f −1 ◦ f = ι.

A.2 Equivalence relationsSo far we’ve used relations primarily to talk about functions. There is another kindof relation, called an equivalence relation, that comes up in the counting problemswith which we began. In Problem 8 with three distinct flavors, it was probablytempting to say there are 12 flavors for the first pint, 11 for the second, and 10 forthe third, so there are 12 · 11 · 10 ways to choose the pints of ice cream. However,once the pints have been chosen, bought, and put into a bag, there is no way to tellwhich is first, which is second and which is third. What we just counted is lists ofthree distinct flavors—one to one functions from the set {1, 2, 3} in to the set of icecream flavors. Two of those lists become equivalent once the ice cream purchaseis made if they list the same ice cream. In other words, two of those lists becomeequivalent (are related) if they list same subset of the set of ice cream flavors. Tovisualize this relation with a digraph, we would need one vertex for each of the12 · 11 · 10 lists. Even with five flavors of ice cream, we would need one vertex foreach of 5 ·4 ·3 = 60 lists. So for now we will work with the easier to draw question ofchoosing three pints of ice cream of different flavors from four flavors of ice cream.

Problem 344. Suppose we have four flavors of ice cream, V(anilla),C(hocolate), S(trawberry) and P(each). Draw the directed graph whosevertices consist of all lists of three distinct flavors of the ice cream, andwhose edges connect two lists if they list the same three flavors. This graphmakes it pretty clear in how many ways we may choose 3 flavors out of four.How many is it?

A.2. Equivalence relations 139

Problem 345.⇒ Now suppose again we are choosing three distinct flavors ofice cream out of four, but instead of putting scoops in a cone or choosingpints, we are going to have the three scoops arranged symmetrically in acircular dish. Similarly to choosing three pints, we can describe a selectionof ice cream in terms of which one goes in the dish first, which one goesin second (say to the right of the first), and which one goes in third (say tothe right of the second scoop, which makes it to the left of the first scoop).But again, two of these lists will sometimes be equivalent. Once they arein the dish, we can’t tell which one went in first. However, there is a subtledifference between putting each flavor in its own small dish and putting allthree flavors in a circle in a larger dish. Think about what makes the listsof flavors equivalent, and draw the directed graph whose vertices consistof all lists of three of the flavors of ice cream and whose edges connect twolists that we cannot tell the difference between as dishes of ice cream. Howmany dishes of ice cream can we distinguish from one another? (h)

Problem 346. Draw the digraph for Problem 38 in the special case wherewe have four people sitting around the table.

In Problems 344, 345, and 346 (as well as Problems 34, 38, and 39) we can beginwith a set of lists, and say when two lists are equivalent as representations of theobjects we are trying to count. In particular, in Problems 344, 345, and 346 youdrew the directed graph for this relation of equivalence. Technically, you shouldhave had an arrow from each vertex (list) to itself. This is what we mean when wesay a relation is reflexive. Whenever you had an arrow from one vertex to a second,you had an arrow back to the first. This is what we mean when we say a relation issymmetric.

When people sit around a round table, each list is equivalent to itself: if List1and List 2 are identical, then everyone has the same person to the right in both lists(including the first person in the list being to the right of the last person). To see thesymmetric property of the equivalence of seating arrangements, if List1 and List2are different, but everyone has the same person to the right when they sit accordingto List2 as when they sit according to List1, then everybody better have the sameperson to the right when they sit according to List1 as when they sit according toList2.

In Problems 344, 345 and 346 there is another property of those relations youmay have noticed from the directed graph. Whenever you had an arrow from L1

to L2 and an arrow from L2 to L3, then there was an arrow from L1 to L3. This iswhat we mean when we say a relation is transitive. You also undoubtedly noticedhow the directed graph divides up into clumps of mutually connected vertices.This is what equivalence relations are all about. Let’s be a bit more precise in ourdescription of what it means for a relation to be reflexive, symmetric or transitive.

• If R is a relation on a set X, we say R is reflexive if (x , x) ∈ R for every x ∈ X.

140 A. Relations

• If R is a relation on a set X, we say R is symmetric if (x , y) is in R whenever(y , x) is in R.

• If R is a relation on a set X, we say R is transitive if whenever (x , y) is in Rand (y , z) is in R, then (x , z) is in R as well.

Each of the relations of equivalence you worked with in Problems 344, 345 and346 had these three properties. Can you visualize the same three properties in therelations of equivalence that you would use in Problems 34, 38, and 39? We call arelation an equivalence relation if it is reflexive, symmetric and transitive.

After some more examples, we will see how to show that equivalence relationshave the kind of clumping property you saw in the directed graphs. In our firstexample, using the notation (a , b) ∈ R to say that a is related to B is going to getin the way. It is really more common to write aRb to mean that a is related to b.For example, if our relation is the less than relation on {1, 2, 3}, you are much morelikely to use x < y than you are (x , y) ∈ <, aren’t you? The reflexive law thensays xRx for every x in X, the symmetric law says that if xRy, then yRx, and thetransitive law says that if xRy and yRz, then xRz.

Problem 347. For the necklace problem, Problem 43, our lists are lists ofbeads. What makes two lists equivalent for the purpose of describing anecklace? Verify explicitly that this relationship of equivalence is reflexive,symmetric, and transitive.

Problem 348. Which of the reflexive, symmetric and transitive propertiesdoes the < relation on the integers have?

Problem 349. A relation R on the set of ordered pairs of positive integersthat you learned about in grade school in another notation is the relationthat says (m , n) is related to (h , k) if mk = hn. Show that this relation is anequivalence relation. In what context did you learn about this relation ingrade school? (h)

Problem 350. Another relation that you may have learned about in school,perhaps in the guise of “clock arithmetic,” is the relation of equivalencemodulo n. For integers (positive, negative, or zero) a and b, we write a ≡ b( n) to mean that a − b is an integer multiple of n, and in this case, wesay that a is congruent to b modulo n and write a ≡ b ( n).. Show thatthe relation of congruence modulo n is an equivalence relation.


Problem 351. Define a relation on the set of all lists of n distinct integerschosen from {1, 2, . . . , n}, by saying two lists are related if they have thesame elements (though perhaps in a different order) in the first k places,and the same elements (though perhaps in a different order) in the last n − kplaces. Show this relation is an equivalence relation.

Problem 352. Suppose that R is an equivalence relation on a set X and foreach x ∈ X, let Cx = {y |y ∈ X and yRx}. If Cx and Cz have an element yin common, what can you conclude about Cx and Cz (besides the fact thatthey have an element in common!)? Be explicit about what property(ies)of equivalence relations justify your answer. Why is every element of X insome set Cx? Be explicit about what property(ies) of equivalence relationsyou are using to answer this question. Notice that we might simultaneouslydenote a set by Cx and Cy . Explain why the union of the sets Cx is X.Explain why two distinct sets Cx and Cz are disjoint. What do these setshave to do with the “clumping” you saw in the digraph of Problem 344 andProblem 345?

In Problem 352 the sets Cx are called equivalence classes of the equivalencerelation R. You have just proved that if R is an equivalence relation of the setX, then each element of X is in exactly one equivalence class of R. Recall that apartition of a set X is a set of disjoint sets whose union is X. For example, {1, 3},{2, 4, 6}, {5} is a partition of the set {1, 2, 3, 4, 5, 6}. Thus another way to describewhat you proved in Problem 352 is the following:

Theorem A.2.1. If R is an equivalence relation on X, then the set of equivalence classes ofR is a partition of X.

Since a partition of S is a set of subsets of S, it is common to call the subsets intowhich we partition S the blocks of the partition so that we don’t find ourselves inthe uncomfortable position of referring to a set and not being sure whether it is theset being partitioned or one of the blocks of the partition.

Problem 353. In each of Problems 38, Problem 39, Problem 43, Problem 344,and Problem 345, what does an equivalence class correspond to? (Fiveanswers are expected here.) (h)

Problem 354. Given the partition {1, 3}, {2, 4, 6}, {5} of the set{1, 2, 3, 4, 5, 6}, define two elements of {1, 2, 3, 4, 5, 6} to be related if theyare in the same part of the partition. That is, define 1 to be related to 3 (and 1and 3 each related to itself), define 2 and 4, 2 and 6, and 4 and 6 to be related(and each of 2, 4, and 6 to be related to itself), and define 5 to be related toitself. Show that this relation is an equivalence relation.

142 A. Relations

Problem 355. Suppose P = {S1 , S2 , S3 , . . . , Sk} is a partition of S. Definetwo elements of S to be related if they are in the same set Si , and otherwisenot to be related. Show that this relation is an equivalence relation. Showthat the equivalence classes of the equivalence relation are the sets Si .

In Problem 353 you just proved that each partition of a set gives rise to anequivalence relation whose classes are just the parts of the partition. Thus inProblem 352 and Problem 353 you proved the following Theorem.

Theorem A.2.2. A relation R is an equivalence relation on a set S if and only if S may bepartitioned into sets S1, S2, . . . , Sn in such a way that x and y are related by R if and onlyif they are in the same block Si of the partition.

In Problems 344, Problem 345, Problem 38 and Problem 43 what we were doingin each case was counting equivalence classes of an equivalence relation. Therewas a special structure to the problems that made this somewhat easier to do. Forexample, in Problem 344, we had 4 · 3 · 2 = 24 lists of three distinct flavors chosenfrom V, C, S, and P. Each list was equivalent to 3 · 2 · 1 = 3! = 6 lists, includingitself, from the point of view of serving 3 small dishes of ice cream. The order inwhich we selected the three flavors was unimportant. Thus the set of all 4 · 3 · 2lists was a union of some number n of equivalence classes, each of size 6. By theproduct principle, if we have a union of n disjoint sets, each of size 6, the unionhas 6n elements. But we already knew that the union was the set of all 24 lists ofthree distinct letters chosen from our four letters. Thus we have 6n = 24, or n = 4equivalence classes.

In Problem 345 there is a subtle change. In the language we adopted for seatingpeople around a round table, if we choose the flavors V, C, and S, and arrange themin the dish with C to the right of V and S to the right of C, then the scoops are indifferent relative positions than if we arrange them instead with S to the right ofV and C to the right of S. Thus the order in which the scoops go into the dish issomewhat important—somewhat, because putting in V first, then C to its right andS to its right is the same as putting in S first, then V to its right and C to its right. Inthis case, each list of three flavors is equivalent to only three lists, including itself,and so if there are n equivalence classes, we have 3n = 24, so there are 24/3 = 8equivalence classes.

Problem 356. If we have an equivalence relation that divides a set with kelements up into equivalence classes each of size m, what is the number nof equivalence classes? Explain why.

Problem 357. In Problem 347, what is the number of equivalence classes?Explain in words the relationship between this problem and the Problem 39.


Problem 358. Describe explicitly what makes two lists of beads equivalentin Problem 43 and how Problem 356 can be used to compute the number ofdifferent necklaces.

Problem 359. What are the equivalence classes (write them out as sets oflists) in Problem 45, and why can’t we use Problem 356 to compute thenumber of equivalence classes?

In Problem 356 you proved our next theorem. In Chapter 1 (Problem 42) wediscovered and stated this theorem in the context of partitions and called it theQuotient Principle

Theorem A.2.3. If an equivalence relation on a set S size k has n equivalence classes eachof size m, then the number of equivalence classes is k/m.

144 A. Relations

Appendix B

Mathematical Induction

B.1 The Principle of Mathematical InductionB.1.1 The ideas behind mathematical inductionThere is a variant of one of the bĳections we used to prove the Pascal Equationthat comes up in counting the subsets of a set. In the next problem it will help uscompute the total number of subsets of a set, regardless of their size. Our maingoal in this problem, however, is to introduce some ideas that will lead us to oneof the most powerful proof techniques in combinatorics (and many other branchesof mathematics), the principle of mathematical induction.

Problem 360.

(a) Write down a list of the subsets of {1, 2}. Don’t forget the empty set!Group the sets containing containing 2 separately from the others.

(b) Write down a list of the subsets of {1, 2, 3}. Group the sets containing3 separately from the others.

(c) Look for a natural way to match up the subsets containing 2 in Part awith those not containing 2. Look for a way to match up the subsetscontaining 3 in Part b containing 3 with those not containing 3.

(d) On the basis of the previous part, you should be able to find a bĳectionbetween the collection of subsets of {1, 2, . . . , n} containing n andthose not containing n. (If you are having difficulty figuring out thebĳection, try rethinking Parts a and b, perhaps by doing a similarexercise with the set {1, 2, 3, 4}.) Describe the bĳection (unless you arevery familiar with the notation of sets, it is probably easier to describeto describe the function in words rather than symbols) and explainwhy it is a bĳection. Explain why the number of subsets of {1, 2, . . . , n}containing n equals the number of subsets of {1, 2, . . . , n − 1}.

145

146 B. Mathematical Induction

(e) Parts a and b suggest strongly that the number of subsets of a n-elementset is 2n . In particular, the empty set has 20 subsets, a one-elementset has 21 subsets, itself and the empty set, and in Parts a and b wesaw that two-element and three-element sets have 22 and 23 subsetsrespectively. So there are certainly some values of n for which ann-element set has 2n subsets. One way to prove that an n-elementset has 2n subsets for all values of n is to argue by contradiction.For this purpose, suppose there is a nonnegative integer n such thatan n-element set doesn’t have exactly 2n subsets. In that case theremay be more than one such n. Choose k to be the smallest such n.Notice that k − 1 is still a positive integer, because k can’t be 0, 1, 2, or3. Since k was the smallest value of n we could choose to make thestatement “An n-element set has 2n subsets” false, what do you knowabout the number of subsets of a (k − 1)-element set? What do youknow about the number of subsets of the k-element set {1, 2, . . . , k}that don’t contain k? What do you know about the number of subsetsof {1, 2, . . . , k} that do contain k? What does the sum principle tellyou about the number of subsets of {1, 2, . . . , k}? Notice that thiscontradicts the way in which we chose k, and the only assumptionthat went into our choice of k was that “there is a nonnegative integern such that an n-element set doesn’t have exactly 2n subsets." Sincethis assumption has led us to a contradiction, it must be false. Whatcan you now conclude about the statement “for every nonnegativeinteger n, an n-element set has exactly 2n subsets?"

Problem 361. The expression

1 + 3 + 5 + · · ·+ 2n − 1

is the sum of the first n odd integers. Experiment a bit with the sum for thefirst few positive integers and guess its value in terms of n. Now apply thetechnique of Problem 360 to prove that you are right. (h)

In Problems 360 and 361 our proofs had several distinct elements. We had astatement involving an integer n. We knew the statement was true for the firstfew nonnegative integers in Problem 360 and for the first few positive integers inProblem 361. We wanted to prove that the statement was true for all nonnegativeintegers in Problem 360 and for all positive integers in Problem 361. In both caseswe used the method of proof by contradiction; for that purpose we assumed thatthere was a value of n for which our formula wasn’t true. We then chose k to bethe smallest value of n for which our formula wasn’t true. This meant that whenn was k − 1, our formula was true, (or else that k − 1 wasn’t a nonnegative integerin Problem 360 or that k − 1 wasn’t a positive integer in Problem 361). What wedid next was the crux of the proof. We showed that the truth of our statement forn = k − 1 implied the truth of our statement for n = k. This gave us a contradiction

B.1. The Principle of Mathematical Induction 147

to the assumption that there was an n that made the statement false. In fact, wewill see that we can bypass entirely the use of proof by contradiction. We used itto help you discover the central ideas of the technique of proof by mathematicalinduction.

The central core of mathematical induction is the proof that the truth of astatement about the integer n for n = k − 1 implies the truth of the statement forn = k. For example, once we know that a set of size 0 has 20 subsets, if we haveproved our implication, we can then conclude that a set of size 1 has 21 subsets,from which we can conclude that a set of size 2 has 22 subsets, from which we canconclude that a set of size 3 has 23 subsets, and so on up to a set of size n having 2n

subsets for any nonnegative integer n we choose. In other words, although it wasthe idea of proof by contradiction that led us to think about such an implication,we can now do without the contradiction at all. What we need to prove a statementabout n by this method is a place to start, that is a value b of n for which weknow the statement to be true, and then a proof that the truth of our statement forn = k − 1 implies the truth of the statement for n = k whenever k > b.

B.1.2 Mathematical inductionThe principle of mathematical induction states that

In order to prove a statement about an integer n, if we can

1. Prove the statement when n = b, for some fixed integer b2. Show that the truth of the statement for n = k −1 implies the truth

of the statement for n = k whenever k > b,

then we can conclude the statement is true for all integers n ≥ b.

As an example, let us return to Problem 360. The statement we wish to prove is thestatement that “A set of size n has 2n subsets.”

Our statement is true when n = 0, because a set of size 0 is the empty setand the empty set has 1 = 20 subsets. (This step of our proof is called abase step.) Now suppose that k > 0 and every set with k − 1 elementshas 2k−1 subsets. Suppose S = {a1 , a2 , . . . ak} is a set with k elements.We partition the subsets of S into two blocks. Block B1 consists of thesubsets that do not contain an and block B2 consists of the subsets thatdo contain an . Each set in B1 is a subset of {a1 , a2 , . . . ak−1}, and eachsubset of {a1 , a2 , . . . ak−1} is in B1. Thus B1 is the set of all subsets of{a1 , a2 , . . . ak−1}. Therefore by our assumption in the first sentence ofthis paragraph, the size of B1 is 2k−1. Consider the function from B2

to B1 which takes a subset of S including an and removes an from it.This function is defined on B2, because every set in B2 contains an . Thisfunction is onto, because if T is a set in B1, then T ∪ {ak} is a set in B2

which the function sends to T. This function is one-to-one because if Vand W are two different sets in B2, then removing ak from them givestwo different sets in B1. Thus we have a bĳection between B1 and B2,so B1 and B2 have the same size. Therefore by the sum principle thesize of B1 ∪ B2 is 2k−1 + 2k−1 = 2k . Therefore S has 2k subsets. This


shows that if a set of size k − 1 has 2k−1 subsets, then a set of size k has2k subsets. Therefore by the principle of mathematical induction, a setof size n has 2n subsets for every nonnegative integer n.

The first sentence of the last paragraph is called the inductive hypothesis. In aninductive proof we always make an inductive hypothesis as part of proving thatthe truth of our statement when n = k − 1 implies the truth of our statement whenn = k. The last paragraph itself is called the inductive step of our proof. In aninductive step we derive the statement for n = k from the statement for n = k − 1,thus proving that the truth of our statement when n = k − 1 implies the truth ofour statement when n = k. The last sentence in the last paragraph is called theinductive conclusion. All inductive proofs should have a base step, an inductivehypothesis, an inductive step, and an inductive conclusion.

There are a couple details worth noticing. First, in this problem, our base stepwas the case n = 0, or in other words, we had b = 0. However, in other proofs, bcould be any integer, positive, negative, or 0. Second, our proof that the truth of ourstatement for n = k − 1 implies the truth of our statement for n = k required thatk be at least 1, so that there would be an element ak we could take away in orderto describe our bĳection. However, condition (2) of the principle of mathematicalinduction only requires that we be able to prove the implication for k > 0, so wewere allowed to assume k > 0.

Problem 362. Use mathematical induction to prove your formula fromProblem 361.

B.1.3 Proving algebraic statements by induction

Problem 363. Use mathematical induction to prove the well-known for-mula that for all positive integers n,

1 + 2 + · · ·+ n =n(n + 1)

2.

Problem 364. Experiment with various values of n in the sum

1

1 · 2 +1

2 · 3 +1

3 · 4 + · · ·+ 1

n · (n + 1)=

n∑i=1

1

i · (i + 1).

Guess a formula for this sum and prove your guess is correct by induction.

Problem 365. For large values of n, which is larger, n2 or 2n? Use mathe-matical induction to prove that you are correct. (h)

B.1. The Principle of Mathematical Induction 149

Problem 366. What is wrong with the following attempt at an inductiveproof that all integers in any consecutive set of n integers are equal for everypositive integer n? For an arbitrary integer i, all integers from i to i are equal,so our statement is true when n = 1. Now suppose k > 1 and all integers inany consecutive set of k −1 integers are equal. Let S be a set of k consecutiveintegers. By the inductive hypothesis, the first k − 1 elements of S are equaland the last k − 1 elements of S are equal. Therefore all the elements in theset S are equal. Thus by the principle of mathematical induction, for everypositive n, every n consecutive integers are equal. (h)

B.1.4 Strong InductionOne way of looking at the principle of mathematical induction is that it tells usthat if we know the “first” case of a theorem and we can derive each other case ofthe theorem from a smaller case, then the theorem is true in all cases. Howeverthe particular way in which we stated the theorem is rather restrictive in thatit requires us to derive each case from the immediately preceding case. Thisrestriction is not necessary, and removing it leads us to a more general statementof the principal of mathematical induction which people often call the strongprinciple of mathematical induction. It states:

In order to prove a statement about an integer n if we can

1. prove our statement when n = b and2. prove that the statements we get with n = b, n = b+1, . . . n = k−1

imply the statement with n = k,

then our statement is true for all integers n ≥ b.

Problem 367. What postage do you think we can make with five and sixcent stamps? Is there a number N such that if n ≥ N , then we can make ncents worth of postage?

You probably see that we can make n cents worth of postage as long as n is at least20. However you didn’t try to make 26 cents in postage by working with 25 cents;rather you saw that you could get 20 cents and then add six cents to that to get26 cents. Thus if we want to prove by induction that we are right that if n ≥ 20,then we can make n cents worth of postage, we are going to have to use the strongversion of the principle of mathematical induction.

We know that we can make 20 cents with four five-cent stamps. Now we let kbe a number greater than 20, and assume that it is possible to make any amountbetween 20 and k − 1 cents in postage with five and six cent stamps. Now if k isless than 25, it is 21, 22, 23, or 24. We can make 21 with three fives and one six.We can make 22 with two fives and two sixes, 23 with one five and three sixes, and24 with four sixes. Otherwise k − 5 is between 20 and k − 1 (inclusive) and so byour inductive hypothesis, we know that k − 5 cents can be made with five and six


cent stamps, so with one more five cent stamp, so can k cents. Thus by the (strong)principle of mathematical induction, we can make n cents in stamps with five andsix cent stamps for each n ≥ 20.

Some people might say that we really had five base cases, n = 20, 21, 22, 23, and24, in the proof above and once we had proved those five consecutive base cases,then we could reduce any other case to one of these base cases by successivelysubtracting 5. That is an appropriate way to look at the proof. A logician wouldsay that it is also the case that, for example, by proving we could make 22 cents, wealso proved that if we can make 20 cents and 21 cents in stamps, then we could alsomake 22 cents. We just didn’t bother to use the assumption that we could make 20cents and 21 cents! So long as one point of view or the other satisfies you, you areready to use this kind of argument in proofs.

Problem 368. A number greater than one is called prime if it has no factorsother than itself and one. Show that each positive number is either a primeor a power of a prime or a product of powers of prime numbers.

Problem 369. Show that the number of prime factors of a positive numbern ≥ 2 is less than or equal to 2 n. (If a prime occurs to the kth power in afactorization of n, you can consider that power as k prime factors.) (Thereis a way to do this by induction and a way to do it without induction. Itwould be ideal to find both ways.)

Problem 370. One of the most powerful statements in elementary numbertheory is Euclid’s Division Theorem.a This states that if m and n are positiveintegers, then there are unique nonnegative intergers q and r with 0 ≤ r < n,such that m = nq + r. The number q is called the quotient and the numberr is called the remainder. In computer science it is common to denote r bym n. In elementary school you learned how to use long division tofind q and r. However, it is unlikely that anyone ever proved for you thatfor any pair of positive intgers, m and n, there is such a pair of nonnegativenumbers q and r. You now have the tools needed to prove this. Do so. (h)

aIn a curious twist of language, mathematicians have long called The Division Algorithm orEuclid’s Division Algorithm. However as computer science has grown in importance, the wordalgorithm has gotten a more precise definition: an algorithm is now a method to do something.There is a method (in fact there are more than one) to get the q and r that Euclid’s DivisionTheorem gives us, and computer scientists would call these methods algorithms. Your authorhas chosen to break with mathematical tradition and restrict his use of the word algorithm tothe more precise interpretation as a computer scientist probably would. We aren’t giving amethod here, so this is why the name used here is “Euclid’s Division Theorem.”

Appendix C

Exponential GeneratingFunctions

C.1 Indicator FunctionsWhen we introduced the idea of a generating function, we said that the formal sum

n∑i=0

ai xi

may be thought of as a convenient way to keep track of the sequence ai . We then didquite a few examples that showed how combinatorial properties of arrangementscounted by the coefficients in a generating function could be mirrored by algebraicproperties of the generating functions themselves. The monomials xi are calledindicator polynomials. (They indicate the position of the coefficient ai .) Oneexample of a generating function is given by

(1 + x)n =∞∑

i=0

(ni

)xi .

Thus we say that (1+ x)n is the generating function for the binomial coefficients(n

i ). The notation tells us that we are assuming that only i varies in the sum on theright, but that the equation holds for each fixed integer n. This is implicit whenwe say that (1 + x)n is the generating function for (n

i ), because we haven’t writteni anywhere in (1 + x)n , so it is free to vary.

Another example of a generating function is given by

xn =∞∑

i=0

s(n , i)xi .

Thus we say that xn is the generating function for the Stirling numbers of thefirst kind, s(n , i). There is a similar equation for Stirling numbers of the secondkind, namely

xn =∞∑

i=0

S(n , i)xi .

151

152 C. Exponential Generating Functions

However with our previous definition of generating functions, this equationwould not give a generating function for the Stirling numbers of the second kind,because S(n , i) is not the coefficient of xi . If we were willing to consider the fallingfactorial powers xi as indicator polynomials, then we could say that xn is thegenerating function for the numbers S(n , i) relative to these indicator polynomials.This suggests that perhaps different sorts of indicator polynomials go naturallywith different sequences of numbers.

The binomial theorem gives us yet another example.

Problem 371.◦ Write (1+x)n as a sum of multiples of xi

i! rather than as a sumof multiples of xi .

This example suggests that we could say that (1+ x)n is the generating functionfor the falling factorial powers ni relative to the indicator polynomials xi

i! . Ingeneral, a sequence of polynomials is called a family of indicator polynomials ifthere is one polynomial of each nonnegative integer degree in the sequence. Thosefamiliar with linear algebra will recognize that this says that a family of indicatorpolynomials form a basis for the vector space of polynomials. This means that eachpolynomial way can be expressed as a sum of numerical multiples of indicatorpolynomials in one and only one way. One could use the language of linear algebrato define indicator polynomials in an even more general way, but a definition insuch generality would not be useful to us at this point.

C.2 Exponential Generating FunctionsWe say that the expression

∑∞i=0 ai

xi

i! is the exponential generating function for thesequence ai . It is standard to use EGF as a shorthand for exponential generatingfunction. In this context we call the generating function

∑ni=0 ai xi that we originally

studied the ordinary generating function for the sequence ai . You can see whywe use the term exponential generating function by thinking about the exponentialgenerating function (EGF) for the all ones sequence,

∞∑i=0

1xi

i!=

∞∑i=0

xi

i!= ex ,

which we also denote by (x). Recall from calculus that the usual definition ofex or (x) involves limits at least implicitly. We work our way around that bydefining ex to be the power series

∑∞i=0

xi

i! .

Problem 372.◦ Find the EGF (exponential generating function) for the se-quence an = 2n . What does this say about the EGF for the number ofsubsets of an n-element set?

C.2. Exponential Generating Functions 153

Problem 373.◦ Find the EGF (exponential generating function) for the num-ber of ways to paint the n streetlight poles that run along the north side ofMain Street in Anytown, USA using four colors.

Problem 374. For what sequence is ex−e−x

2 = x the EGF (exponentialgenerating function)?

Problem 375.· For what sequence is ( 11−x ) the EGF? ( (y) stands for the

natural logarithm of y. People often write (y) instead.) Hint: Think ofthe definition of the logarithm as an integral, and don’t worry at this stagewhether or not the usual laws of calculus apply, just use them as if they do!We will then define (1 − x) to be the power series you get. a

a It is possible to define the derivatives and integrals of power series by the formulas

ddx

∞∑i=0

bi xi =∞∑

i=1

ibi xi−1

and ∫ x

0

∞∑i=0

bi xi =∞∑

i=0

bii + 1

xi+1

rather than by using the limit definitions from calculus. It is then possible to prove that thesum rule, product rule, etc. apply. (There is a little technicality involving the meaning ofcomposition for power series that turns into a technicality involving the chain rule, but itneedn’t concern us at this time.)

Problem 376.· What is the EGF for the number of permutations of an n-element set?

Problem 377.⇒ · What is the EGF for the number of ways to arrange n peoplearound a round table? Try to find a recognizable function represented bythe EGF. Notice that we may think of this as the EGF for the number ofpermutations on n elements that are cycles. (h)

Problem 378.⇒ · What is the EGF∑∞

n=0 p2nx2n

(2n)! for the number of ways p2n topair up 2n people to play a total of n tennis matches (as in Problems 12 and44)? (h)


Problem 379.◦ What is the EGF for the sequence 0, 1, 2, 3, . . .? You may thinkof this as the EFG for the number of ways to select one element from an nelement set. What is the EGF for the number of ways to select two elementsfrom an n-element set?

Problem 380.· What is the EGF for the sequence 1, 1, . . . , 1, . . .? Notice thatwe may think of this as the EGF for the number of identity permutationson an n-element set, which is the same as the number of permutations ofn elements that are products of 1-cycles, or as the EGF for the number ofways to select an n-element set (or, if you prefer, an empty set) from an n-element set. As you may have guessed, there are many other combinatorialinterpretations we could give to this EGF.

Problem 381.◦ What is the EGF for the number of ways to select n distinctelements from a one-element set? What is the EGF for the number of waysto select a positive number n of elements from a one element set? Hint:When you get the answer you will either say “of course,” or “this is a sillyproblem.” (h)

Problem 382.· What is the EGF for the number of partitions of a k-elementset into exactly one block? (Hint: is there a partition of the empty set intoexactly one block?)

Problem 383.· What is the EGF for the number of ways to arrange k bookson one shelf (assuming they all fit)? What is the EGF for the number ofways to arrange k books on a fixed number n of shelves, assuming that allthe books can fit on any one shelf? (Remember Problem 122.)

C.3 Applications to recurrences.We saw that ordinary generating functions often play a role in solving recurrencerelations. We found them most useful in the constant coefficient case. Exponentialgenerating functions are useful in solving recurrence relations where the coeffi-cients involve simple functions of n, because the n! in the denominator can cancelout factors of n in the numerator.

C.3. Applications to recurrences. 155

Problem 384.◦ Consider the recurrence an = nan−1+n(n−1). Multiply bothsides by xn

n! , and sum from n = 2 to ∞. (Why do we sum from n = 2 toinfinity instead of n = 1 or n = 0?) Letting y =

∑∞i=0 ai xi , show that the

left-hand side of the equation is y − a0 − a1x. Express the right hand side interms of y, x, and ex . Solve the resulting equation for y and use the resultto get an equation for an . (A finite summation is acceptable in your answerfor an .)

Problem 385.⇒ · The telephone company in a city has n subscribers. Assumea telephone call involves exactly two subscribers (that is, there are no calls tooutside the network and no conference calls), and that the configuration ofthe telephone network is determined by which pairs of subscribers are talk-ing. Notice that we may think of a configuration of the telephone networkas a permutation whose cycle decomposition consists entirely of one-cyclesand two-cycles, that is, we may think of a configuration as an involution inthe symmetric group Sn .

(a) Give a recurrence for the number cn of configurations of the network.(Hint: Person n is either on the phone or not.)

(b) What are c0 and c1?

(c) What are c2 through c6?

Problem 386.⇒ · Recall that a derangement of [n] is a permutation of [n] thathas no fixed points, or equivalently is a way to pass out n hats to their ndifferent owners so that nobody gets the correct hat. Use dn to stand forthe number of derangements of [n]. We can think of derangement of [n] asa list of 1 through n so that i is not in the ith place for any n. Thus in aderangement, some number k different from n is in position n. Considertwo cases: either n is in position k or it is not. Notice that in the second case,if we erase position n and replace n by k, we get a derangement of [n − 1].Based on these two cases, find a recurrence for dn . What is d1? What is d2?What is d0? What are d3 through d6?

C.3.1 Using calculus with exponential generating functions

Problem 387.⇒ · Your recurrence in Problem 385 should be a second orderrecurrence.

(a) Assuming that the left hand side is cn and the right hand side involvescn−1 and cn−2, decide on an appropriate power of x divided by an ap-


propriate factorial by which to multiply both sides of the recurrence.Using the fact that the derivative of xn

n! is xn−1(n−1)! , write down a differen-

tial equation for the EGF T(x) =∑∞

i=0 cixi

i! . Note that it makes senseto substitute 0 for x in T(x). What is T(0)? Solve your differentialequation to find an equation for T(x).

(b) Use your EGF to compute a formula for cn . (h)

Problem 388.⇒ · Your recurrence in Problem 386 should be a second orderrecurrence.

(a) Assuming that the left-hand side is dn and the right hand side involvesdn−1 and dn−2, decide on an appropriate power of x divided by an ap-propriate factorial by which to multiply both sides of the recurrence.Using the fact that the derivative of xn

n! is xn−1(n−1)! , write down a differen-

tial equation for the EGF D(x) =∑∞

i=0 dixi

i! . What is D(0)? Solve yourdifferential equation to find an equation for D(x).

(b) Use the equation you found for D(x) to find an equation for dn . Com-pare this result with the one you computed by inclusion and exclusion.

C.4 The Product Principle for EGFsOne of our major tools for ordinary generating functions was the product principle.It is thus natural to ask if there is a product principle for exponential generatingfunctions. In Problem 383 you likely found that the EGF for the number of waysof arranging n books on one shelf was exactly the same as the EGF for the numberof permutations of [n], namely 1

1−x or (1 − x)−1. Then using our formula fromProblem 122 and the generating function for multisets, you probably found thatthe EGF for number of ways of arranging n books on some fixed number m ofbookshelves was (1− x)−m . Thus the EGF for m shelves is a product of m copies ofthe EGF for one shelf.

Problem 389.◦ In Problem 373 what would the exponential generating func-tion have been if we had asked for the number of ways to paint the poleswith just one color of paint? With two colors of paint? What is the rela-tionship between the EGF for painting the n poles with one color of paintand the EGF for painting the n poles with four colors of paint? What is therelationship between the EGF for painting the n poles with two colors ofpaint and the EGF for painting the poles with four colors of paint?

In Problem 385 you likely found that the EGF for the number of network con-figurations with n customers was ex+x2/2 = ex · ex2/2. In Problem 380 you saw

C.4. The Product Principle for EGFs 157

that the generating function for the number of permutations on n elements thatare products of one cycles was ex , and in Problem 378 you likely found that theEGF for the number of tennis pairings of 2n people, or equivalently, the number ofpermutations of 2n objects that are products of n two-cycles is ex2/2.

Problem 390.· What can you say about the relationship among the EGF forthe number of permutations that are products of disjoint two-cycles andone-cycles, i.e., are involutions, the exponential generating function for thenumber of permutations that are the product of disjoint two-cycles onlyand the generating function for the number of permutations that are theproduct of disjoint one cycles only (these are identity permutations on theirdomain)?

In Problem 388 you likely found that the EGF for the number of permutationsof [n] that are derangements is e−x

1−x . But every permutation is a product of derange-ments and one cycles, because the permutation that sends i to i is a one-cycle, sothat when you factor a permutation as a product of disjoint cycles, the cycles ofsize greater than one multiply together to give a derangement, and the elementsnot moved by the permutation are one-cycles.

Problem 391.· If we multiply the EGF for derangements times the EGF forthe number of permutations whose cycle decompositions consist of one-cycles only, what EGF do we get? for what set of objects have we found theEGF? (h)

We now have four examples in which the EGF for a sequence or a pair of objectsis the product of the EGFs for the individual objects making up the sequence orpair.

Problem 392.· What is the coefficient of xn

n! in the product of two EGFs∑∞i=0 ai

xi

i! and∑∞

j=0 b jx j

j! ? (A summation sign is appropriate in your an-swer.) (h)

In the case of painting streetlight poles in Problem 389, let us examine therelationship between the EGF for painting poles with two colors, the EGF forpainting the poles with three colors, and the EGF for painting poles with fivecolors, e5x . To be specific, the EGF for painting poles red and white is e2x and theEGF for painting poles blue, green, and yellow is e3x . To decide how to paint poleswith red, white, blue, green, and yellow, we can decide which set of poles is tobe painted with red and white, and which set of poles is to be painted with blue,green, and yellow. Notice that the number of ways to paint a set of poles with redand white depends only on the size of that set, and the number of ways to paint aset of poles with blue, green, and yellow depends only on the size of that set.


Problem 393.· Suppose that ai is the number of ways to paint a set of i poleswith red and white, and b j is the number of ways to paint a set of j poleswith blue, green, and yellow. In how many ways may we take a set N of npoles, divide it up into two sets I and J (using i to stand for the size of I andj to stand for the size of the set J, and allowing i and j to vary) and paint thepoles in I red and white and the poles in J blue, green, and yellow? (Giveyour answer in terms of ai and b j . Don’t figure out formulas for ai and b j touse in your answer; that will make it harder to get the point of the problem!)How does this relate to Problem 392?

Problem 393 shows that the formula you got for the coefficient of xn

n! in theproduct of two EGFs is the formula we get by splitting a set N of poles intotwo parts and painting the poles in the first part with red and white and thepoles in the second part with blue, green, and yellow. More generally, you couldinterpret your result in Problem 392 to say that the coefficient of xn

n! in the product∑∞i=0 ai

xi

i!∑∞

j=0 b jx j

j! of two EGFs is the sum, over all ways of splitting a set N of sizen into an ordered pair of disjoint sets I and J, of the product a |I |b | J | .

There seem to be two essential features that relate to the product of exponentialgenerating functions. First, we are considering structures that consist of a set andsome additional mathematical construction on or relationship among the elementsof that set. For example, our set might be a set of light poles and the additionalconstruction might be a coloring function defined on that set. Other examples ofadditional mathematical constructions or relationships on a set could include apermutation of that set; in particular an involution or a derangement, a partitionof that set, a graph on that set, a connected graph on that set, an arrangement ofthe elements of that set around a circle, or an arrangement of the elements of thatset on the shelves of a bookcase. In fact a set with no additional construction orarrangement on it is also an example of a structure. Its additional construction is theempty set! When a structure consists of the set S plus the additional construction,we say the structure uses S. What all the examples we have mentioned in our earlierdiscussion of exponential generating functions have in common is that the numberof structures that use a given set is determined by the size of that set. We will call afamily F of structures a species of structures on subsets of a set X if structures aredefined on finite subsets of X and if the number of structures in the family using afinite set S is finite and is determined by the size of S (that is, if there is a bĳectionbetween subsets S and T of X, the number of structures in the family that use Sequals the number of structures in the family that use T). We say a structure is anF -structure if it is a member of the family F .

Problem 394.· In Problem 383, why is the family of arrangements of set ofbooks on a single shelf (assuming they all fit) a species?


Problem 395.· In Problem 385, why is the family of people actually makingphone calls (assuming nobody is calling outside the telephone network) atany given time, with the added relationship of who is calling whome, aspecies? Why is the family of sets of people who are not using their phonesa species (with no additional construction needed)?

The second essential feature of our examples of products of EGFs is that prod-ucts of EGFs seem to count structures on ordered pairs of two disjoint sets (or moregenerally on k-tuples of mutually disjoint sets). For example, we can determinea five coloring of a set S by partitioning it in all possible ways into two sets andcoloring the first set in the pair with our first two colors and our second pair withthe last three colors. Or we can partition our set in all possible ways into five partsand color part i with our ith color. We don’t have to do the same thing to eachpart of our partition; for example, we could define a derangement on one part andan identity permutation on the other; this defines a permutation on the set we arepartitioning, and we have already noted that every permutation arises in this way.

Our combinatorial interpretation of EGFs will involve assuming that the coeffi-cient of xi

i! counts the number of structures on a particular set of of size i in a speciesof structures on subsets of a set X. Thus in order to give an interpretation of theproduct of two EGFs we need to be able to think of ordered pairs of structures ondisjoint sets or k-tuples of structures on disjoint sets as structures themselves. Thusgiven a structure on a set S and another structure on a disjoint set T, we define theordered pair of structures (which is a mathematical construction!) to be a structureon the set S ∪T. We call this a pair structure on S ∪T. We can get many structureson a set S ∪ T in this way, because S ∪ T can be divided into many other pairs ofdisjoint sets. In particular, the set of pair structures whose first structure comesfrom F and whose second element comes from G is denoted by F · G.

Problem 396. Show that if F and G are species of structures on subsets ofa set X, then the pair of structures of F · G for a species of structures

Given a species F of structures, the number of structures using any particularset of size i is the same as the number of structures in the family using any otherset of size i. We can thus define the exponential generating function (EGF) for thefamily as the power series

∑∞i=1 ai

xi

i! , where ai is the number of structures of F thatuse one particular set of size i. In Problems 372, 373, 376, 377, 378, 380, 382, 383,387, and 388, we were computaing EGFs for species of subsets of some set.

Problem 397. If F and G are species of subsets of X, how is the EGF forF · G related to the EGFs for F and G? Prove you are right. (h)


Problem 398. Without giving the proof, how can you compute the EGF f (x)for the number of structures using a set of size n in the species F1 · F2 · · · · · Fkof structures on k-tuples of subsets of X from the EGFs fi(x) for Fi for eachi from 1 to k? (Here we are using the natural extension of the idea of thepair of structures to the idea of a k-tuple structure.)

Theorem C.4.1. If F1 , F2 , . . . , Fn are species of subsets of the set X and Fi has EGF fi(x),then the family of k-tuple structures F1 · F2 · · · · · Fn has EGF

∏ni=1 fi(x).

We call Theorem C.4.1 the product principle for exponential generating func-tions. We give two corollaries; the proof of the second is not immediate thoughnot particular difficult.

Corollary C.4.2. If F is a species of structures on subsets of X and f (x0) is the EGFfor F , then f (x)k is the EGF for the k-tuple structure on k-tuples of F -structures usingdisjoint subsets of X.

Our next corollary uses the idea of a k-set structure. Suppose we have a speciesF of structures on nonempty subsets of X, that is, a species of structures whichassigns no structures to the empty set. Then we can define a new species F (k) ofstructures, called “k-set structures,” using nonempty subsets of X. Given a fixedpositive integer k, a k-set structure on a subset Y of X consists of a k-elementset of nonempty disjoint subsets of X whose union is Y and an assignment of anF -structure to each of the disjoint subsets. This is a species on the set of subsets ofX; the subset used by a k-set structure is the union of the sets of the structure. Torecapitulate, the set of k-set structures on a subset Y of X is the set of all possibleassignments of F -structures to k nonempty disjoint sets whose union is Y. (Youcan also think of the k-set structures as a family of structures defined on blocks ofpartitions of subsets of X into k blocks.)

Corollary C.4.3. If F is a species of structures on nonempty subsets of X and f (x) is theEGF for F , then for each positive integer k, f (x)k

k! is the EGF for the family F (k) of k-setstructures on subsets of X

Problem 399. Prove Corollary C.4.3. (h)

Problem 400.· Use the product principle for EGFs to explain the results ofProblems 390 and Problem 391.

Problem 401.· Use the general product principle for EGFs or one of its corol-laries to explain the relationship between the EGF for painting streetlightpoles in only one color and the EGF for painting streetlight poles in 4 colors


in Problems 373 and Problem 389. What is the EGF for the number pn ofways to paint n streetlight poles with some fixed number k of colors of paint.

Problem 402.· Use the general product principle for EGFs or one of its corol-laries to explain the relationship between the EGF for arranging books onone shelf and the EGF for arranging books on n shelves in Problem 383.

Problem 403.⇒ (Optional) Our very first example of exponential generatingfunctions used the binomial theorem to show that the EGF for k-elementpermutations of an n element set is (1+ x)n . Use the EGF for k-element per-mutations of a one-element set and the product principle to prove the samething. Hint: Review the alternate definition of a function in Section 3.1.2. (h)

Problem 404. What is the EGF for the number of ways to paint n streetlightpoles red, white blue, green and yellow, assuming an even number of polesmust be painted green and an even number of poles must be painted yellow?Give a formula for the number of ways to paint n poles. (Don’t forget thefactorial!) (h)

Problem 405.⇒ · What is the EGF for the number of functions from an n-element set onto a one-element set? (Can there be any functions from theempty set onto a one-element set?) What is the EGF for the number cn offunctions from an n-element set onto a k element set (where k is fixed)? Usethis EGF to find an explicit expression for the number of functions from ak-element set onto an n-element set and compare the result with what yougot by inclusion and exclusion.

Problem 406.⇒ · In Problem 142 You showed that the Bell Numbers Bn satisfythe equation Bn+1 =

∑nk=0 (

nk )Bn−k (or a similar equation for Bn .) Multiply

both sides of this equation by xn

n! and sum from n = 0 to infinity. On the lefthand side you have a derivative of a certain EGF we might call B(x). Onthe right hand side, you have a product of two EGFs, one of which is B(x).What is the other one? What differential equation involving B(x) does thisgive you. Solve the differential equation for B(x). This is the EGF for theBell numbers!.


Problem 407.⇒ Prove that n2n−1 =∑n

k=1 (nk )k by using EGFs. (h)

Problem 408.· In light of Problem 382, why is the EGF for the Stirling num-bers S(n , k) of the second kind not (ex − 1)n? What is it equal to instead?

C.5 The Exponential FormulaExponential generating functions turn out to be quite useful in advanced work incombinatorics. One reason why is that it is often possible to give a combinatorialinterpretation to the composition of two exponential generating functions. Inparticular, if f (x) =

∑ni=0 ai

xi

i! and g(x) =∑∞

j=1 b jx j

j! , it makes sense to form thecomposition f (g(x)) because in so doing we need add together only finitely manyterms in order to find the coefficient of xn

n! in f (g(x)) since in the EGF g(x) thedummy variable j starts at 1. Since our study of combinatorial structures hasnot been advanced enough to give us applications of a general formula for thecomposition of the EGF, we will not give here the combinatorial interpretation ofthis composition. However we have seen some examples where one particularcomposition can be applied. Namely, if f (x) = ex = (x), then f (g(x)) =exp(g(x)) is well defined when b0 = 0. We have seen three examples in which anEGF is e f (x) where f (x) is another EGF. There is a fourth example in which theexponential function is slightly hidden.

Problem 409.· If f (x) is the EGF for the number of partitions of an n-setinto one block, and g(x) is the EGF for the total number of partitions of ann-element set, that is, for the Bell numbers Bn , how are the two generatingfunctions related?

Problem 410.· Let f (x) be the EGF for the number of permutations of ann-element set with one cycle of size one or two. What is f (x)? What is theEGF g(x) for the number of permutations of an n-element set all of whosecycles have size one or two, that is, the number of involutions in Sn? Howare these two exponential generating functions related?

Problem 411.⇒ · Let f (x) be the EGF for the number of permutations of an n-element set that have exactly one two-cycle and no other cycles. Let g(x) bethe EGF for the number of permutations which are products of two-cyclesonly, that is, for tennis pairings. (That is, they are not a product of two-cycles

C.5. The Exponential Formula 163

and a nonzero number of one-cycles). What is f (x)? What is g(x)? Howare these to exponential generating functions related?

Problem 412.· Let f (x) be the EGF for the number of permutations of ann-element set that have exactly one cycle. (This is the same as the EGF forthe number of ways to arrange n people around a round table.) Let g(x) bethe EGF for the total number of permutations of an n-element set. What isf (x)? What is g(x)? How are f (x) and g(x) related?

There was one element that our last four problems had in common. In eachcase our EGF f (x) involved the number of structures of a certain type (partitions,telephone networks, tennis pairings, permutations) that used only one set of anappropriate kind. (That is, we had a partition with one part, a telephone networkconsisting either of one person or two people connected to each other, a tennispairing of one set of two people, or a permutation with one cycle.) Our EGF g(x)was the number of structures of the same “type” (we put type in quotation markshere because we don’t plan to define it formally) that could consist of any numberof sets of the appropriate kind. Notice that the order of these sets was irrelevant.For example we don’t order the blocks of a partition and a product of disjoint cyclesis the same no matter what order we use to write down the product. Thus we wererelating the EGF for structures which were somehow “building blocks” to the EGFfor structures which were sets of building blocks. For a reason that you will seelater, it is common to call the building blocks connected structures. Notice thatour connected structures were all based on nonempty sets, so we had no connectedstructures whose value was the empty set. Thus in each case, if f (x) =

∑∞i=0 ai

xi

i! ,we would have a0 = 0. The relationship between the EGFs was always g(x) = e f (x).We now give a combinatorial explanation for this relationship.

Problem 413.· Suppose that F is a species of structures of a set X with nostructures on the empty set. Let f (x) be the EGF for F .

(a) In the power series

e f (x) = 1 + f (x) +f (x)2

2!+ · · ·+ f (x)k

k!+ · · · =

∞∑k=0

f (x)k

k!,

what does Corollary C.4.3 tell us about the coefficient of xn

n! in f (x)k

k! ?

(b) What does the coefficient of xn

n! in e f (x) count?

In Problem 413 we proved the following theorem, which is called the exponen-tial formula.


Theorem C.5.1. Suppose that F is a species of structures on subsets of a set X with nostructures on the empty set. Let f (x) be the EGF for F . Then the coefficient of xn

n! in e f (x)

is the number of sets of structures on disjoint sets whose union is a particular set of size n.

Let us see how the exponential formula applies to the examples in Problems 409,410, 411 and 412. In Problem 382 our familyF should consist of one-block partitionsof finite subsets of a set, say the set of natural numbers. Since a partition of a setis a set of blocks whose union is S, a one-block partition whose block is B is theset {B}. Then any nonempty finite subset of of the positive integers is the valueof exactly one structure in F . (There is no one-block partition of the empty set,so we have no structures using the empty set.) As you showed in Problem 382the generating function for partitions with just one block is ex − 1. Thus by theexponential formula, (ex−1) is the EGF for sets of subsets of the positive integerswhose values are disjoint sets whose union is any particular set N of size n. Thisset of disjoint sets partitions the set N . Thus (ex − 1) is the EGF for partitions ofsets of size n. (As we wrote our description, it is the EGF for partitions of n-elementsubsets of the positive integers, but any two n-element sets have the same numberof partitions.) In other words, (ex − 1) is the exponential generating functionfor the Bell numbers Bn .

Problem 414.· Explain how the exponential formula proves the relationshipwe saw in Problem 412.



Problem 417.· In Problem 373 we saw that the generating function for thenumber of ways to use five colors of paint to paint n light poles alongthe north side of Main Street in Anytown was e4x . We should expect anexplanation of this EGF using the exponential formula. Let F be the familyof all one-element sets of light poles with the additional construction ofan ordered pair consisting of a light pole and a color. Thus a given lightpole occurs in five ordered pairs. Put no structures on any other finite set.Show that this is a species of structures on the finite subsets of the positiveintegers. What is the exponential generating function f (x) forF ? Assumingthat there is no upper limit on the number of light poles, what subsets of Sdoes the exponential formula tell us are counted by the coefficient of xn ine f (x)? How do the sets being counted relate to ways to paint light poles?

C.5. The Exponential Formula 165

One of the most spectacular applications of the exponential formula is alsothe reason why, when we regard a combinatorial structure as a set of buildingblock structures, we call the building block structures connected. In Chapter 2 weintroduced the idea of a connected graph and in Problem 104 we saw examples ofgraphs which were connected and were not connected. A subset C of the vertex setof a graph is called a connected component of the graph if

• every vertex in C is connected to every other vertex in that set by a walkwhose vertices lie in C, and

• no other vertex in the graph is connected by a walk to any vertex in C.

In Problem 241 we showed that each connected component of a graph consistsof a vertex and all vertices connected to it by walks in the graph.

Problem 418.· Show that every vertex of a graph lies in one and only oneconnected component of a graph. (Notice that this shows that the connectedcomponents of a graph form a partition of the vertex set of the graph.)

Problem 419.· Explain why no edge of the graph connects two vertices indifferent connected components.

Problem 420.· Explain why it is that if C is a connected component of agraph and E′ is the set of all edges of the graph that connect vertices in C,then the graph with vertex set C and edge set E′ is a connected graph. Wecall this graph a connected component graph of the original graph.

The last sequence of problems shows that we may think of any graph as the setof its connected component graphs. (Once we know them, we know all the verticesand all the edges of the graph). Notice that a graph is connected if and only ifit has exactly one connected component. Since the connected components form apartition of the vertex set of a graph, the exponential formula will relate the EGFfor the number of connected graphs on n vertices with the EGF for the number ofgraphs (connected or not) on n vertices. However because we can draw as manyedges as we want between two vertices of a graph, there are infinitely many graphson n vertices, and so the problem of counting them is uninteresting. We can makeit interesting by considering simple graphs, namely graphs in which each edgehas two distinct endpoints and no two edges connect the same two vertices. It isbecause connected graphs form the building blocks for viewing all graphs as sets ofconnected components that we refer to the building blocks for structures countedby the EGF in the exponential formula as connected structures.


Problem 421.⇒ · Suppose that f (x) =∑∞

n=0 cnxn

n! is the exponential generatingfunction for the number of simple connected graphs on n vertices andg(x) =

∑∞i=0 ai

xi

i! is the exponential generating function for the number ofsimple graphs on i vertices. From this point onward, any use of the wordgraph means simple graph.

(a) Is f (x) = e g(x), is f (x) = e g(x)−1, is g(x) = e f (x)−1 or is g(x) = e f (x)? (h)

(b) One of ai and cn can be computed by recognizing that a simple graphon a vertex set V is completely determined by its edge set and its edgeset is a subset of the set of two element subsets of V . Figure out whichit is and compute it. (h)

(c) Write g(x) in terms of the natural logarithm of f (x) or f (x) in termsof the natural logarithm of g(x).

(d) Write (1 + y) as a power series in y. (h)

(e) Why is the coefficient of x0

0! in g(x) equal to one? Write f (x) as a powerseries in g(x) − 1.

(f) You can now use the previous parts of the problem to find a formulafor cn that involves summing over all partitions of the integer n. (Itisn’t the simplest formula in the world, and it isn’t the easiest formulain the world to figure out, but it is nonetheless a formula with whichone could actually make computations!) Find such a formula. (h)

The point to the last problem is that we can use the exponential formula inreverse to say that if g(x) is the generating function for the number of (nonempty)connected structures of size n in a given family of combinatorial structures andf (x) is the generating function for all the structures of size n in that family, thennot only is f (x) = e g(x), but g(x) = ( f (x)) as well. Further, if we happen to havea formula for either the coefficients of f (x) or the coefficients of g(x), we can get aformula for the coefficients of the other one!

C.6 Supplementary Problems1. Use product principle for EGFs and the idea of coloring a set in two colors toprove the formula ex · ex = e2x .

2. Find the EGF for the number of ordered functions from a k-element set to ann-element set.

3. Find the EGF for the number of ways to string n distinct beads onto a necklace.

4. Find the exponential generating function for the number of broken permutationsof a k-element set into n parts.

5. Find the EGF for the total number of broken permutations of a k-element set.

C.6. Supplementary Problems 167

6. Find the EGF for the number of graphs on n vertices in which every vertex hasdegree 2.

7. Recall that a cycle of a permutation cannot be empty.(a) What is the generating function for the number of cycles on an even number

of elements (i.e. permutations of an even number n of elements that forman n-cycle)? Your answer should not have a summation sign in it. Hint: Ify =

∑∞i=0

x2i

2i , what is the derivative of y?(b) What is the generating function for the number of permutations on n elements

that are a product of even cycles?(c) What is the generating function for the number of cycles on an odd number

of elements?(d) What is the generating function for the number of permutations on n elements

that are a product of odd cycles?(e) How do the generating functions in parts b and d of this problem related to

the generating function for all permutations on n elements?

Appendix D

Hints to Selected Problems

1. Answer the questions in Problem 2 for the case of five schools.

3. For each kind of bread, how many sandwiches are possible?

6. Try to solve the problem first with a two-scoop cone. (Look for an earlierproblem that is analogous.) Then, for each two scoop cone, in how manyways can you put on a top scoop?

7.a. Ask yourself “how many choices do we have for f (1)?” Then ask how manychoices we have for f (2).

7.b. It may not be practical to write down rules for all the functions for thisproblem. But you could ask yourself how many choices we have for f (1),how many we have for f (2) and how many we have for f (3).

7.c. If you are choosing a function f , how many choices do you have for f (a)?Then how many choices do you have for f (b)?

8.a. You know how to figure out in how many ways they could make a list ofthree flavors out of the twelve. But each set of three flavors can be listed in anumber of different ways. Try to figure out in how many ways a set of threeflavors can be listed, and then try to see how this helps you.

8.b. Try to break the problem up into cases you can solve by previous methods;then figure out how to get the answer to the problem by using these answersfor the cases.

12.a. Suppose you have a list in alphabetical order of names of the members of theclub. In how many ways can you pair up the first person on the list? In howmany ways can you pair up the next person who isn’t already paired up?

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170 D. Hints to Selected Problems

15. In how many ways may you assign the men to their rows? The women? Oncea woman and a man have a row to share, in how many ways may they choosetheir seats?

18. Try applying the product principle in the case n = 2 and n = 3. How mightyou apply it in general?

19. Ask yourself if either the sum principle or product principle applies.Additional Hint: Remember that zero is a number.

20. Do you see an analogy between this problem and any of the previous prob-lems?

26.a. For each part of this problem, think about how many arrows are allowed toenter a vertex representing a member of Y.

28. The problem is asking you to describe a one-to-one function from the set ofbinary representations of numbers between 0 and 2n−1 onto the set of subsetsof the set [n]. Write down these two sets for n = 2. They should both havefour elements. The set of binary representations should contain the string 00.You could think of this as the instruction “take no ones and take no twos.”In that context, what could you think of the string 11 as standing for? Thisshould help you describe a function. Of course now you have to figure outhow to show it is one-to-one and onto.

31. Starting with the row 1 8 28 56 70 56 28 8 1, put dots below it where theelements of row 9 should be. Then put dots below that where the elementsof row 10 should be. Do the same for rows 11 and 12. Mark the dot whererow 12 should appear. Now mark the dots you need in row 11 to computethe entry in column 3 of row 12. Now mark the dots you need in row 10 tocompute the marked entries in row 11. Do the same for rows 9 and 8. Nowyou should be able to see what you need to do.

32.a. Begin by trying to figure out what the entries just above the diagonal of therectangle are. After that, what other entries can you figure out?

32.b. See if you can figure out what the entries in column −1 have to be.

32.c. What does the sum of two consecutive values in row −1 have to be? Couldthis sum depend on which two consecutive values we take? Is there somevalue of row −1 that we could choose arbitrarily? Now what about row −2?Can we make arbitrary choices there? If so, how many can we make, and istheir position arbitrary?

36. The first thing you need to decide is “What are the two sets whose elementswe are counting?” Then it will be easier to think of a bĳection between thesetwo sets. It turns out that these two sets are sets of sets!

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37. Ask yourself “What is a problem like this doing in the middle of a bunch ofproblems about counting subsets of a set? Is it related, or is it supposed togives us a break from sets?”

38. The problem suggests that you think about how to get a list from a seatingarrangement. Could every list of n distinct people come from a seating chart?How many lists of n distinct people are there? How many lists could we getfrom a given seating chart by taking different starting places?

Additional Hint: For a different way of doing the problem, suppose that youhave chosen one person, say the first one in a list of the people in alphabeticalorder by name. Now seat that person. Does it matter where they sit? In wayscan you seat the remaining people? Does it matter where the second personin alphabetical order sits?

39.a. A block consists of all permutations of some subset {a1 , a2 , . . . , ak} of S. Howmany permutations are there of the set {a1 , a2 , . . . , ak}?

39.c. What sets are listed, and how many times is each one listed if you take onelist from each row of Table 1.2.8? How does this choice of lists give you thebĳection in this special case?

39.d. You can make good use of the product principle here.

40.b. The coach is making a sequence of decisions. Can you figure out how manychoices the coach has for each decision in the sequence?

40.c. As with any counting problem whose context does not suggest an approach,it is useful to ask yourself if you could decompose the problem into simplerparts by using either the sum or product principle.

43. How could we get a list of beads from a necklace?

Additional Hint: When we cut the necklace and string it out on a table, thereare 2n lists of beads we could get. Why is it 2n rather than n?

44.a. You might first choose the pairs of people. You might also choose to make alist of all the people and then take them by twos from the list.

44.b. You might first choose ordered pairs of people, and have the first person ineach pair serve first. You might also choose to make a list of all the peopleand then take them by twos from the list in order.

45. It might be helpful to just draw some pictures of the possible configurations.There aren’t that many.

47. Note that we must walk at least ten blocks, so ten is the smallest number ofblocks possible. In how many of those ten blocks must we walk north?


48.b. In Problem 47 you saw that we had to make ten choices of north or east,choosing north four times.

48.c. This problem is actually a bit tricky. What happens to the answer if i > m orj > n? Remember that paths go up or to the right.

49.a. Where can you go from (0,0) in one step? In two steps? In any of these cases,what can you say about the sum of the coordinates of a point you can get to?Can you find any other relationship between the x- and y-coordinates of apoint you can get to? For example, can you get to the point (1, 3)?

49.c. How many choices do you have to make in order to choose a path?

50.c. In each part, each such sequence corresponds to a path that can’t cross over(but may touch) a certain line.

51.b. Given a path from (0, 0) to (n , n) which touches or crosses the line y = x +1,how can you modify the part of the path from (0, 0) to the first touch ofy = x + 1 so that the modified path starts instead at (−1, 1)? The trick is todo this in a systematic way that will give you your bĳection.

51.c. A path either touches the line y = x + 1 or it doesn’t. This partitions the setof paths into two blocks.

52.b. Look back at the definition of a Dyck Path and a Catalan Path.

52.c. What makes this part difficult is understanding how we are partitioning thepaths. As an example, B0 is the set of all paths that have no upsteps followingthe last absolute minimum. Can such a path have downsteps after the lastabsolute minimum? (The description we gave of B0 is not succinct enough tobe the answer to the second question of this part.) As another example B1 isthe set of all paths that have exactly one upstep and perhaps some downstepsafter the last absolute minimum. Is it possible, though, for a path in B1 to haveany downsteps after the last absolute minimum? A path in B2 has exactly twoupsteps after its last absolute minimum. If is possible to have one downstepafter the last absolute minimum, but it has to be in a special place. Whatplace is that? Now to figure out how many parts our partition has, we needto know the maximum number of upsteps a path can have following its lastabsolute minimum. What is this maximum? It might help to draw somepictures with n = 5 or 6. In particular, is it possible that all upsteps occurafter the last absolute minimum?

52.e. Using d for down and u for up, we could have uudduuddudud as our Catalanpath. Suppose that i = 5. The fifth upstep is the u in position 9. ThusF = uudduudd, U = u, and B = dud. Now BUF is duduuudduudd. Thisis a Dyck path that begins by going below the x-axis. The d’s in positions 1and 3 take the path to the y-coordinate −1. Then the y coordinate climbs to 2,

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goes back to 0, up to 2 again, and finally down to 0. So the absolute minimumis −1, and it occurs in the first and third position. There are five u’s afterthe third positon. So this Dyck path is in the block B5 of our partition. Nowcomes the crucial question. Why were there five u’s after that last absoluteminimum in position 3? Try with the same path and i = 3. Figure out whythere are three u’s after the last absolute minimum in the resulting path. Allthis discussion should explain why when i = 5, the set of all Catalan pathsis mapped into the set B5. Keeping i = 5 for a while, try to see why thiscorrespondence between Catalan paths and B5 is a bĳection. Then, if youneed to, do the same thing with i = 3. This should give you enough insightto do the general case.

55. What would the lower limit of the sum have to be for this problem to be aroutine application of the binomial theorem?

56. What does the binomial theorem give you for (x − y)n?

57. Consider (x + y)m(x + y)n .

Additional Hint: What does (m+nk ) count? What does (m

i )(m

k−1 ) count?

58. For example when n = 3, we have (30 ) = (33 ) and (31 ) = (32 ). The numberof subsets of even size is (30 ) + (32 ) and the number of subsets of odd size is(31 ) + (33 ), and the two sums can be paired off into equal terms. When wesubtract the number of subsets of odd size from the number of subsets ofeven size, the pairing also gives us (30 ) − (31 ) + (32 ) − (33 ) = 0.

59. Take the derivative of something interesting.

61. To prove that each function from a set S of size n to a set of size less thann is not one-to-one, we must prove that regardless of the function f that wechoose, there are always two elements, say x and y, such that f (x) = f (y).

62. The previous exercise could help you prove that if f is one-to-one, then it isonto.

Additional Hint: The sum principle can help you show that if f is an ontofunction, then f is one-to-one.

63. The statement of the generalized pigeonhole principle involves the numberof elements in a block, so a counting principle is likely to help you.

64. You may choose a specific number for n if you want to. Notice that the lasttwo digits of the powers of a prime other than two cannot represent an evennumber.

65. While this sounds like a pigeonhole principle problem, the ordinary pigeon-hole principle doesn’t guarantee three of something.


67. What usually makes it hard for students to start this problem is the fact thatwe just defined what R(4, 4) is, and not what it means for a number not to beR(4, 4). So to get started, try to write down what it means to say R(4, 4) isnot 8. You will see that there are two things that can keep R(4, 4) from being8. You need to figure out which one happens and explain why. One suchexplanation could involve the graph K8 .

68. Review Problem 65 and your solution of it.

69. Let ai be the number of acquaintances of person i. Can you explain why thesum of the numbers ai is even?

70. Often when there is a counter-example, there is one with a good deal ofsymmetry. (Caution: there is a difference between often and always!) Oneway to help yourself get a symmetric example, if there is one, is to put 8vertices into a circle. Then, perhaps, you might draw green edges in somesort of regular fashion until it is impossible to draw another green edgebetween any two of the vertices without creating a green triangle.

71. In Problem 68 you showed that R(4, 3) ≤ 10. In Problem 70 you showed thatR(4, 3) > 8. Thus R(4, 3) is either 9 or 10. Deciding which is the case is justplain hard. But there is a relevant problem we have done that we haven’tused yet.

72. We wish to prove that (ni ) = n!

i!(n−i)! . Mathematical induction allows us toassume that (n−1

j ) =(n−1)!

j!(n−1− j)! for every jbetween 0 and n − 1. How does thisput us into a position to use the Pascal relation? What special cases will beleft over?

73. What sort of relationship do you know between values of the form (ni ) and

values of the form (n−1j )?

75. We did something rather similar in our example of the inductive proof that aset with n elements has 2n subsets. The work you did in a previous problemmay be similar to part of what you need to do here.

76.a. This may look difficult because one can’t decide in advance on whether totry to induct on m, on n, or on their sum. In some sense, it doesn’t matterwhich you choose to induct on, though inducting on the sum would lookmore complicated. For most people inducting on n fits their way of workingwith exponents best.

76.b. Here it matters whether you choose to induct on m or n. However, it mattersonly in the sense that you need to use more tools in one case. In one case, youare likely to need the rule (cd)n = cn dn(, which we haven’t proved. (However,you might be able to prove that by induction!) In either case, you may findpart (a) handy.

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79. We didn’t explicitly say to use induction here, but, especially in this context,induction is a natural tool to try here. But we don’t have a variable n toinduct on. That means you have to choose one. So what do you think is mostuseful. The number of blocks in the partition? The size of the first block ofthe partition? The size of the set we are partitioning? Or something else?

80. Think about how you might have gone from the number of double deckercones to the number of triple decker cones in Problem 6.

81. Perhaps the first thing one needs to ask is why proving that if there are(m+n−2

m−1 )people in a room, then there are either at least m mutual acquaintancesor at least n mutual strangers proves that R(m , n) exists. Can you see whythis tells us that there is some number R of people such that if R people are ina room, then there are m mutual acquaintances or n mutual strangers? Andwhy does that mean the Ramsey Number exists?

Additional Hint: Naturally it should come as no surprise that you will usedouble induction, and you can use either form. As you think about how touse induction, the Pascal relation will come to mind. This suggests that youwant to make assumptions involving (m+n−3

m−1 ) people in a room, or (m+n−3m−2 )

people in a room. Now you have to figure out what these assumptions areand how they help you prove the result! Recall that we have made progressbefore by choosing one person and asking whether this person is acquaintedwith at least some number of people or unacquainted with at least some othernumber of people.

82. One expects to need double induction again here. But only because of thelocation of the problem and because the sum looks like double induction.And those reasons aren’t enough to mean you have to use double induction.If you had this result in hand already, then you could us it with doubleinduction to give a second proof that Ramsey Numbers exist.

Additional Hint: What you do need to show is that if there are R(m − 1, n)+R(m , n−1) people in a room, then there are either m mutual acquaintances orn mutual strangers. As with earlier problems, it helps to start with a personand think about the number of people with whom this person is acquaintedor nonacquainted. The generalized pigeonhole principle tells you somethingabout these numbers.

83.b. If you could find four mutual acquaintances, you could assume person 1 isamong them. And by the generalized pigeonhole principle and symmetry,so are two of the people to the first, second, fourth and eighth to the right.Now there are lots of possibilities for that fourth person. You now have thehard work of using symmetry and the definition of who is acquainted withwhom to eliminate all possible combinations of four people. Then you haveto think about nonacquaintances.

86.a. What is the definition of R(n , n)?


86.b. If you average a bunch of numbers and each one is bigger than one, what canyou say about the average?

86.c. Note that there are 2(n2) graphs on a set of n vertices.

86.d. A notation for the sum over all colorings c of Km is∑c:c is a coloring of Km

,

and a notation for the sum over all subsets N of M that have size n is∑N:N⊆M, |N |=n

86.e. If you interchange the order of summation so that you sum over subsets firstand colorings second, you can take advantage of the fact that for a fixed subsetN , you can count count the number of colorings in which it is monochromatic.

86.f. You have an inequality involving m and n that tells you that R(n , n) > m.Suppose you could work with that inequality in order to show that if theinequality holds, then m is bigger than something. What could you concludeabout R(n , n)?

87. Remember, a subset of [n] either does or doesn’t contain n.

90.b. A first order recurrence for an gives us an as a function of an−1.

91. Suppose you already knew the number of moves needed to solve the puzzlewith n − 1rings.

92. If we have n − 1circles drawn in such a way that they define rn−1 regions, andwe draw a new circle, each time it crosses another circle, except for the lasttime, it finishes dividing one region into two parts and starts dividing a newregion into two parts.

Additional Hint: Compare rn with the number of subsets of an n-elementset.

98. You might try working out the cases n = 2, 3, 4 and then look for a pattern.Alternately, you could write an−1 = ban−2 + d, substitute the right hand sideof this expression into an = ban−1 + d to get a recurrence involving only an−2, and then repeat a similar process with an−2 and perhaps an−3 and see apattern that is developing.

102.a. There are several ways to see how to do this problem. One is to draw picturesof graphs with one edge, two edges, three edges, perhaps four edges and fig-ure out the sum of the degrees. Another is to ask what deleting an edge does

177

to the sum of the degrees. Another is to ask what a given edge “contributes”to the sum of the degrees.

102.b. To make your inductive step, think about what happens to a graph if youdelete an edge.

102.d. Suppose that instead of summing the degree of v over all vertices v, you sumsome quantity defined for each edge e over all the edges.

103. Whatever you say should be consistent with what you already know aboutdegrees of vertices.

108. What happens if you choose an edge and delete it, but not its endpoints?

109. One approach to the problem is to use facts that we already know aboutdegrees, vertices and edges. Another approach is to try deleting an edgefrom a tree with more than one vertex and analyze the possible numbers ofvertices of degree one in what is left over.

111. When you get to four and especially five vertices, draw all the unlabeled treesyou can think of, and then figure out in how many different ways you can putlabels on the vertices.

112.b. Do some examples.

112.c. Is it possible for a1 to be equal to one of the b js?

112.d. You have seen that the sequence b determines a1. Does it determine any othera js? If you knew all the a js and all the b js, could you reconstruct the tree?What are the possible values of b1? b j?

113. What vertex or vertices in the sequence b1 , b2 , . . . , bn−1 can have degree 1?

115. If a vertex has degree 1, how many times does it appear in the Prüfer code ofthe tree? What about a vertex of degree 2?

116. How many vertices appear exactly once in the Prüfer code of the tree andhow many appear exactly twice?

118. Think of selecting one edge of the tree at a time. Given that you have chosensome edges and have a graph whose connected components are trees, whatis a good way to choose the next edge? To prove your method correct, usecontradiction by assuming there is a spanning tree tree with lower total cost.

Additional Hint: Think of selecting one edge of the tree at a time. But nowdo it in such a way that one connected component is a tree and the otherconnected components have just one vertex. What is a good way to make


the component that is a tree into a tree with one more vertex? To prove yourmethod works, use contradiction by assuming there is a spanning tree withlower total cost.

119.a. If you have a spanning tree of G that contains e, is the graph that results fromthat tree by contracting e still a tree?

122.c. If you decide to put it on a shelf that already has a book, you have two choicesof where to put it on that shelf.

122.e. Among all the places you could put books, on all the shelves, how many areto the immediate left of some book? How many other places are there?

123. How can you make sure that each shelf gets at least one book before you startthe process described in Problem 122?

124. What is the relationship between the number of ways to distribute identicalbooks and the number of ways to distribute distinct books?

125. Look for a relationship between a multiset of shelves and a way of distributingidentical books to shelves

126. Note that (n+k−1k ) = (n+k−1

n−1 ). So we have to figure out how choosing either kelements or n − 1 elements out of n + k − 1 elements constitutes the choice ofa multiset. We really have no idea what set of n + k − 1 objects to use, so whynot use [n + k −1]? If we choose n −1 of these objects, there are k left over, thesame number as the number of elements of our multiset. Since our multisetis supposed to be chosen from an n-element set, perhaps we should let then-element set be [n]. From our choice of n − 1 numbers, we have to decideon the multiplicity of 1 through n. For example with n = 4 and k = 6, wehave n + k − 1 = 9. Here, shown with underlines, is a selection of 3 = n − 1elements from [9]: 1, 2, 3, 45, 6, 7, 8, 9. How do the underlined elements giveus a multiset of size 6 chosen from an [4]-element set? In this case, 1 hasmultiplicity 2, 2 has multiplicity 1, 3 has multiplicity 2, and 4 has multiplicity1.

127. A solution to the equations assigns a nonnegative number to each of 1, 2, . . . ,mso that the nonnegative numbers add to r. Does such an assignment have acombinatorial meaning?

128. Can you think of some way of guaranteeing that each recipient gets m objects(assuming k ≥ mn) right at the beginning of the process of passing the objectsout?

129. We already know how to place k distinct books onto n distinct shelves sothat each shelf gets at least one. Suppose we replace the distinct books with

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identical ones. If we permute the distinct books before replacement, doesthat affect the final outcome? There are other ways to solve this problem.

130. Do you see a relationship between compositions and something else we havecounted already?

131. If we line up k identical books, how many adjacencies are there in betweenbooks?

133. Imagine taking a stack of k books, and breaking it up into stacks to put intothe boxes in the same order they were originally stacked. If you are goingto use n boxes, in how many places will you have to break the stack up intosmaller stacks, and how many ways can you do this?

Additional Hint: How many different bookcase arrangements correspond tothe same way of stacking k books into n boxes so that each box has at leastone book?

134. The number of partitions of [k] into n parts in which k is not in a block relatesto the number of partitions of k − 1 into some number of blocks in a way thatinvolves n. With this in mind, review how you proved Pascal’s (recurrence)equation.

137. What if the question asked about six sandwiches and two distinct bags? Howdoes having identical bags change the answer?

138. What are the possible sizes of parts?

139. Suppose we make a list of the k items. We take the first k1 elements to be theblocks of size 1. How many elements do we need to take to get k2 blocks ofsize two? Which elements does it make sense to choose for this purpose?

141. To see how many broken permutations of a k element set into n parts donot have k is a part by itself, ask yourself how many broken permutationsof [7] result from adding 7 to the one of the two permutations in the brokenpermutation {14, 2356}.

142.b. Here it is helpful to think about what happens if you delete the entire blockcontaining k rather than thinking about whether k is in a block by itself ornot.

143. You can think of a function as assigning values to the blocks of its partition.If you permute the values assigned to the blocks, do you always change thefunction?

144. The Prüfer code of a labeled tree is a sequence of n − 2 entries that mustbe chose from the vertices that do not have degree 1. The sequence can be


though of as a function from the set [n − 2] to the set of vertices that do nothave degree 1. What is special about this function?

145. When you add the number of functions mapping onto J over all possiblesubsets J of N , what is the set of functions whose size you are computing?

148. What if the ji ’s don’t add to k?

Additional Hint: Think about listing the elements of the k-element set andlabeling the first j1 elements with label number 1.

149. The sum principle will help here.

150. How are the relevant ji ’s in the multinomial coefficients you use here differentfrom the ji ’s in the previous problem.

151. Think about how binomial coefficients relate to expanding a power of a bino-mial and note that the binomial coefficient (n

k ) and the multinomial coefficient( n

k ,n−k ) are the same.

152.a. We have related Stirling numbers to powers nk . How are binomial coefficientsrelated to falling factorial powers?

152.b. In the equation∑n

j=0 n jS(k , j) = nk , we might try substituting x for n. How-ever we don’t know what

∑xj=0 means when x is a variable. Is there anything

other than n that makes a suitable upper limit for the sum? (Think aboutwhat you know about S(k , j).

153. For the last question, you might try taking advantage of the fact that x =x + 1 − 1.

154. What does induction have to do with Equation (3.1)?

Additional Hint: What could you assume inductively about xk−1 if you weretrying to prove xk =

∑kn=0 s(k , n)xn?

156.a. There is a solution for this problem similar to the solution to Problem 154.

156.b. Is the recurrence you got familiar?

156.d. Show that (−x)k = (−1)k xk and (−x)k = (−1)k xk .

Additional Hint: The first hint lets you write an equation for (−1)k xk as a ris-ing factorial of something else and then use what you know about expressingrising factorials in terms of falling factorials, after which you have to convertback to factorial powers of x.

181

162. How can you start with a partition of k and make it into a new partition ofk+1 that is guaranteed to have a part of size one, even if the original partitiondidn’t?

163. Draw a line through the top-left corner and bottom-right corner of the topleftbox.

164. The largest part of a partition is the maximum number of boxes in a row ofits Young diagram. What does the maximum number of boxes in a columntell us?

165. Draw all self conjugate partitions of integers less than or equal to 8. Drawall partitions of integers less than or equal to 8 into distinct odd parts (manyof these will have just one part). Now try to see how to get from one set ofdrawings to the other in a consistent way.

166. Draw the partitions of six into even parts. Draw the partitions of six intoparts used an even number of times. Look for a relationship between one setof diagrams and the other set of diagrams. If you have trouble, repeat theprocess using 8 or even 10 in place of 6.

167. Draw a partition of ten into four parts. Assume each square has area one.Then draw a rectangle of area 40 enclosing your diagram that touches thetop of your diagram, the left side of your diagram and the bottom of yourdiagram. How does this rectangle give you a partition of 30 into four parts?

168.c. Consider two cases, m′ > m and m′ = m.

168.d. Consider two cases, n′ > n and n′ = n.

169. Suppose we take two repetitions of this complementation process. What rowsand columns do we remove from the diagram?Additional Hint: To deal with an odd number of repetitions of the comple-mentation process, think of it as an even number plus 1. Thus ask what kindof partition gives us the partition of one into one part after this complemen-tation process.

170. How many compositions are there of k into n parts? What is the maximumnumber of compositions that could correspond to a given partition of k inton parts?

171.a. These two operations do rather different things to the number of parts, andyou can describe exactly what only one of the operations does. Think aboutthe Young diagram.

171.b. Think about the Young diagram. In only one of the two cases can you givean exact answer to the question.


171.c. Here the harder part requires that, after removal, you consider a range ofpossible numbers being partitioned and that you give an upper bound on thepart size. However it lets you describe the number of parts exactly.

171.d. One of the two sets of partitions of smaller numbers from the previous partis more amenable to finding a recurrence than the other. The resulting recur-rence does not have just two terms though.

171.h. If there is a sum equal to zero, there may very well be a partition of zero.

172. How does the number of compositions of k into n distinct parts compare tothe number of compositions of k into n parts (not necessarily distinct)? Whatdo compositions have to do with partitions?

173. While you could simply display partitions of 7 into three parts and partitionsof 10 into three parts, we hope you won’t. Perhaps you could write down thepartitions of 4 into two parts and the partitions of 5 into two distinct partsand look for a natural bĳection between them. So the hope is that you willdiscover a bĳection from the set of partitions of 7 into three parts and thepartitions of 10 into three distinct parts. It could help to draw the Youngdiagrams of partitions of 4 into two parts and the partitions of 5 into twodistinct parts.

174. In the case k = 4 and n = 2, we have m = 5. In the case k = 7 and n = 3, wehave m = 10.

175. What can you do to a Young diagram for a partition of k into n distinct partsto get a Young diagram of a partition of k − n into some number of distinctparts?

176. For any partition of k into parts λ1, λ2, etc. we can get a partition of k intoodd parts by factoring the highest power of two that we can from each λi ,writing λi = γi · 2k

i . Why is γi odd? Now partition k into 2k1 parts of size γ1,2k2 parts of size γ2, etc. and you have a partition of k into odd parts.

177. Suppose we have a partition of k into distinct parts. If the smallest part, say m,is smaller than the number of parts, we may add one to each of the m largestparts and delete the smallest part, and we have changed the parity of thenumber of parts, but we still have distinct parts. On the other hand, supposethe smallest part, again say m, is larger than or equal to the number of parts.Then we can subtract 1 from each part larger than m, and add a part equalto the number of parts larger than m. This changes the parity of the numberof parts, but if the second smallest part is m + 1, the resulting partition doesnot have distinct parts. Thus this method does not work. Further, if it didalways work, the case k " 3 j2+ j

2 would be covered also. However you canmodify this method by comparing m not to the total number of parts, but tothe number of rows at the top of the Young diagram that differ by exactly one

183

from the row above. Even in this situation, there are certain slight additionalassumptions you need to make, so this hint leaves you a lot of work to do. (Itis reasonable to expect problems because of that exceptional case.) However,it should lead you in a useful direction.

183. Substitute something for A, P and B in your formula from Problem 181.

184. For example, to get the cost of the fruit selection APB you would want to getx20x25x30 = x75.

186. Consider the example with n = 2. Then we have two variables, x1 and x2 .Forgetting about x2 , what sum says we either take x1 or we don’t? Forgettingabout x1 , what sum says we either take x2 or we don’t? Now what productsays we either take x1 or we don’t and we either take x2 or we don’t?

188. For the last two questions, try multiplying out something simpler first, say(a0+a1x+a2x2)(b0+b1x+b2x2) . If this problem seems difficult the part thatseems to cause students the most problems is converting the expression theyget for a product like this into summation notation. If you are having thiskind of problem, expand the product (a0 + a1x + a2x2)(b0 + b1x + b2x2) andthen figure out what the coefficient of x2 is. Try to write that in summationnotation.

189. Write down the formulas for the coefficients of x0, x1, x2 and x3 in(

n∑i=0

ai xi

) -./

m∑j=0

b j x j012

.

190. How is this problem different from Problem 189? Is this an important differ-ence from the point of view of the coefficient of xk?

191. If this problem appears difficult, the most likely reason is because the defi-nitions are all new and symbolic. Focus on what it means for

∑∞k=0 ck xk to

be the generating function for ordered pairs of total value k. In particular,how do we get an ordered pair with total value k? What do we need to knowabout the values of the components of the ordered pair?

192.b. You might try applying the product principle for generating functions to anappropriate power of the generating function you got in the first part of thisproblem.

Additional Hint: In Problem 125 you found a formula for the number ofk-element multisets chosen from an n-element set. Suppose you use thisformula for ak in

∑∞k=0 ak xk . What do you get the generating function for?

195. While you could use calculus techniques, there is a much simpler approach.Note that 1 + x = 1 − (−x).


Additional Hint: Can you see a way to use Problem 194?

197. Look for a power of a polynomial to get started.

Additional Hint: The polynomial referred to in the first hint is a quotient oftwo polynomials. The power of the denominator can be written as a powerseries.

198. Intepret Problem 197 in terms of multisets.

199.e. When you factor out x1x2 · · · xn from the enumerator of trees, the result is asum of terms of degree n − 2. (The degree of xi1

1 xi22 · · · xin

n is i1 + i2 + · · ·+ in .)

Additional Hint: Write down the picture (using xs) of a tree on five verticeswith two vertices of degree one, of one with three vertices of degree one, andwith four vertices of degree 1. Factor x1x2x3x4x5 out of the picture and lookat what is left. How is it related to your vertices of degree one? Now removethe vertices of degree 1 from the tree and write down the picture of the treethat remains. What is special about the vertices of degree 1 of that tree. (Youcan just barely learn something from this with five vertex trees, so you mightwant to experiment a bit with six or seven vertex trees.)

200. This is a good place to apply the product principle for picture enumerators.

201.a. The product principle for generating functions helps you break the generatingfunction into a product of ten simpler ones.

201.b. m was 10 in the previous part of this problem.

202. Think about conjugate partitions.

203.a. Don’t be afraid of writing down a product of infinitely many power series.

203.b. From the fifth factor on, there is no way to choose a qi that has i nonzero andless than five from the factor.

203.d. Describe to yourself how to get the coefficient of a given power of q.

204. If infinitely many of the polynomials had a nonzero coefficient for q, wouldthe product make any sense?

205. (1+ q2+ q4)(1+ q3+ q9) is the generating function for partitions of an integerinto at most two twos and at most two threes.

206. (1+ q2+ q4)(1+ q3+ q9) is the generating function for partitions of an integerinto at most two twos and at most two threes. (This is intentionally the samehint as in the previous problem, but it has a different point in this problem.)

185

207. In the power series∑∞

j=0 q2i j , the 2i j has a different interpretation if you thinkof it as (2i) · j or if you think of it as i · (2 j).

208. Note that1 − q2

1 − q· 1 − q4

1 − q2· 1 − q6

1 − q3· 1 − q8

1 − q4=

(1 − q6)(1 − q8)

(1 − q)(1 − q3).

209. Note that qi + q3i + q5i + · · · = qi(1 + q2 + q4 + · · · ).

210.a. We want to calculate the number of partitions whose Young diagrams fit intoa two by two square. These partitions have at most two parts and the partshave size at most two. Thus they are partitions of 1, 2, 3, or 4. However notall partitions of 3 or 4 have diagrams that fit into a two by two square. Trywriting down the relevant diagrams.

210.b. They are the generating function for the number of partitions whose Youngdiagram fits into a rectangle n−1units wide and 1 unit deep or into a rectangle1 unit wide and n − 1 units deep respectively.

210.c. How can you get a diagram of a partition counted by partition is counted by[m+n

n ]q from one whose partition is counted by [m+nm ]q?

210.e.iii. Think about geometric operations on Young Diagrams

210.f. How would you use the Pascal recurrence to prove the corresponding resultfor binomial coefficients?

210.g. For finding a bĳection, think about lattice paths.

210.h. If you could prove [m+nn ]q is a polynomial function of q, what would that tell

you about how to compute the limit as q approaches −1?

Additional Hint: Try computing a table of values of [m+nn ]q with q = −1 by

using the recurrence relation. Make a pretty big table so you can see what ishappening.

211.c. You may run into a product of the form∑∞

i=0 ai xi ∑∞j=0 b j x j . Note that in the

product, the coefficient of xk is∑k

i=0 ai bk−i =∑k

i=0ai

bi .

214. Our recurrence becomes an = an−1 + an−2.

217.5x + 1

(x − 3)(x − 5)=

cx + 5c + dx − 3d(x − 3)(x − 5)

gives us

5x = cx + dx


1 = 5c − 3d.

218. To haveax + b

(x − r1)(x − r2)=

cx − r1

+d

x − r2we must have

cx − r2c + dx − r1d = ax + b.

221. You can save yourself a tremendous amount of frustrating algebra if youarbitrarily choose one of the solutions and call it r1 and call the other solutionr2 and solve the problem using these algebraic symbols in place of the actualroots.1 Not only will you save yourself some work, but you will get a formulayou could use in other problems. When you are done, substitute in the actualvalues of the solutions and simplify.

222.a. Once again it will save a lot of tedious algebra if you use the symbols r1 andr2 for the solutions as in Problem 221 and substitute the actual values of thesolutions once you have a formula for an in terms of r1 and r2.

222.d. Think about how the binomial theorem might help you.

224.a. A Catalan path could touch the x-axis several times before it reaches (2n , 0).Its first touch can be any point (2i , 0) between (2, 0) and (2n , 0). For the pathto touch first at (2i , 0), the path must start with an upstep and then proceed asa Dyck path from (1, 1) to (2i−1, 1). From there it must take a downstep. Canyou see a bĳection between such Dyck paths and Catalan paths of a certainkind?

224.b. Does the right-hand side of the recurrence remind you of some products youhave worked with?

224.c.1 · 3 · 5 · · · (2i − 3)

i!=

(2i − 2)!

(i − 1)!2i i!.

226. Try drawing a Venn Diagram.

228. Try drawing a Venn Diagram.

231.b. For each student, how big is the set of backpack distributions in which thatstudent gets the correct backpack? It might be a good idea to first considercases with n = 3, 4, and 5.

Additional Hint: For each pair of students (say Mary and Jim, for example)how big is the set of backpack distributions in which the students in this pairget the correct backpack. What does the question have to do with unions or

1We use the words roots and solutions interchangeably.

187

intersections of sets. Keep on increasing the number of students for whichyou ask this kind of question.

232. Try induction.

Additional Hint: We can apply the formula of Problem 226 to getAAAAA

n⋃i=1

Ai

AAAAA =AAAAA(

n−1⋃i=1

Ai

)∪ An

AAAAA=

AAAAAn−1⋃i=1

Ai

AAAAA+ |An | −AAAAA(

n−1⋃i=1

Ai

)∩ An

AAAAA=

AAAAAn−1⋃i=1

Ai

AAAAA+ |An | −AAAAAn−1⋃i=1

Ai ∩ An

AAAAA233.b. Let T be the set of all i such that x ∈ Ai . In terms of x, what is different about

the i in T and those not in T?

Additional Hint: You may come to a point where the binomial theoremwould be helpful.

235. Notice that it is straightforward to figure out how many ways we may passout the apples so that child i gets five or more apples: give five apples tochild i and then pass out the remaining apples however you choose. And ifwe want to figure out how many ways we may pass out the apples so that agiven set C of children each get five or more apples, we give five to each childin C and then pass out the remaining k − 5|C | apples however we choose.

236. Start with two questions that can apply to any inclusion-exclusion problem.Do you think you would be better off trying to compute the size of a unionof sets or the size of a complement of a union of sets? What kinds of sets(that are conceivably of use to you) is it easy to compute the size of? (Thesecond question can be interpreted in different ways, and for each way ofinterpreting it, the answer may help you see something you can use in solvingthe problem.)

Additional Hint: Suppose we have a set S of couples whom we want to seatside by side. We can think of lining up |S | couples and 2n − 2|S | individualpeople in a circle. In how many ways can we arrange this many items in acircle?

237. Reason somewhat as you did in Problem 236, noting that if the set of coupleswho do sit side-by-side is nonempty, then the sex of the person at each placeat the table is determined once we seat one couple in that set.

Additional Hint: Think in terms of the sets Ai of arrangements of people inwhich couple i sits side-by-side. What does the union of the sets Ai have todo with the problem?


239. What does Problem 238 have to do with this question?

242.a. For each edge in F to connect two vertices of the same color, we must haveall the vertices in a connected component of the graph with vertex set V andedge set F colored the same color.

242.c. How does the number you are trying to compute relate to the union of thesets Ai?

243. One way to get a proper coloring of G − e is to start with a proper coloring ofG and remove e. But there are other colorings of G that become proper whenyou remove e.

246. One approach would be to try to guess the result by doing a bunch of examplesand use induction to prove you are right. If you try this, what will you beable to use to make the induction step work? There are other approaches aswell.

253.a. What do you want ϕn ◦ ϕ−1 to be?

254. If σi = σ j and i " j, what can you conclude about ι?

256.b. What does it mean for one function to be the inverse of another one?

261. Once you know where the corners of the square go under the action of anisometry, how much do you know about the isometry?

264. In how many ways can you choose a place to which you can move vertex1? Having done that, in how many ways can you place the three verticesadjacent to vertex 1?

265.a. In how many ways can you choose a place to which you can move vertex1? Having done that, in how many ways can you place the three verticesadjacent to vertex 1?

265.b. Why is it sufficient to focus on permutations that take vertex 1 to itself?

270. If a subgroup contains, say, ρ3 and some flip, how many elements of D4 mustit contain?

272. If the list (i σ(i) σ2(i) . . . σn(i)) does not have repeated elements but thelist (i σ(i) σ2(i) . . . σn(i) σn+1(i)) does have repeated elements, then whichelement or elements are repeats?

277. The element k is either in a cycle by itself or it isn’t.

189

286. Before you try to show that σ actually is a permutation of the colorings, itwould be useful to verify the second part of the definition of a group action,namely that σ ◦ ϕ = σ ◦ ϕ.

289. If z ∈ Gx and z ∈ Gy , how can you use elements of G to explain therelationship between x and y?

Additional Hint: Suppose σ is a fixed member of G. As τ ranges over G,which elements of G occur as τσ?

295. How does the size of a multiorbit compare to the size of G?

301. We are asking for the number of orbits of some group on lists of four Rs, sixBs, and seven Gs.

305. There are five kinds of elements in the rotation group of the cube. Forexample, there are six rotations by 90 degrees or 270 degrees around an axisconnecting the centers of two opposite faces and there are 8 rotations (of120 degrees and 240 degrees, respectively) around an axis connecting twodiagonally opposite vertices.

306. Is it possible for a nontrivial rotation to fix any coloring?

309. There are 48 elements in the group of automorphisms of the graph.

Additional Hint: For this problem, it may be easier to ask which groupelements fix a coloring rather than which colorings are fixed by a groupelement.

326. The group of automorphisms of the graph has 48 elements and contains D6

as a subgraph.

Additional Hint: The permutations with four one-cycles and the two-cycle(1 4), (2 5), or (3 6) are in the group of automorphisms. Once you know thecycle structure of D6 and (1 4)D6 = {(1 4)σ |σ ∈ D6}, you know the cyclestructure of every element of the group.

327. What does the symmetric group on five vertices have to do with this problem?

329.c. In the relation of a function, how many pairs (x , f (x)) have the same x-value?

332. For the second question, how many arrows have to leave the empty set? Howmany arrows have to leave a set of size one?

339. What is the domain of g ◦ f ?

345. If we have scoops of vanilla, chocolate, and strawberry sitting in a circle in adish, can we distinguish between VCS and VSC?


349. To show a relation is an equivalence relation, you need to show it satisfies thedefinition of an equivalence relation.

353. To get you started, in Problem 38 the equivalence classes correspond to seatingarrangements.

361. You’ve probably guessed that the sum is n2. To prove this by contradiction,you have to assume it is false, that is, that there is an n such that 1 + 3 +5 + · · · + 2n − 1 " n2. Then the method of Problem 360 says there mustbe a smallest such n and suggests we call it k. Why do you know that1 + 3 + 5 + · · · + 2k − 3 = (k − 1)2? What happens if you add 2n − 1 to bothsides of the equation?

365. You’ve probably already seen that, with small values of n, sometimes n2 andsometimes 2n is bigger. But if you keep experimenting one of the functionsseems to get bigger and stay bigger than the other. The number n = b wherethis change occurs is a good choice for a base case. So as not to spoil theproblem for you, we won’t say here what this value of b is. However youshouldn’t be surprised later in the proof if you need to use the assumptionthat n/gtb.Additional Hint: You may have reached the point of assuming that 2k−1 >(k − 1)2 and found yourself wondering how to prove that 2k > k2. A naturalthing to try is multiplying both sides of 2k−1 > (k − 1)2 by 2. This ends upgiving you 2k > 2k2 − 4k + 2. Based on previous experience it is natural foryou to expect to see how to turn this new right hand side into k2 but not seehow to do it. Here is the hint. You only need to show that the right hand sideis greater than or equal to k2. For this purpose you need to show that one ofthe two k2s in 2k2 somehow balances out the −4k. See if you can figure outhow the fact that you are only considering ks with k > b can help you out.

366. When you suspect an argument is not valid, it may be helpful to explicitly tryseveral values of n to see if it makes sense for them. Often small values of nare adequate to find the flaw. If you find one flaw, it invalidates everythingthat comes afterwards (unless, of course, you can fix the flaw).

370. You might start out by ignoring the word unique and give a proof of thesimpler theorem that results. Then look at your proof to see how you caninclude uniqueness in it.

377. An earlier problem may help you put your answer into a simpler form.

378. What is the power series representation of ex2?

381. There is only one element that you may choose. In the first case you eitherchoose it or you don’t.

387.b. At some point, you may find the binomial theorem to be useful.

191

391. Notice that any permutation is a product of a derangement of the elementsnot fixed by the permutation times a permutation whose cycle decompositionconsists of one-cycles.

392. A binomial coefficient is likely to appear in your answer.

397. If f (x) =∑∞

i=0 aixi

i! and g(x) =∑∞

j=0 b jx j

j! , what is the coefficient of xn

n! inf (x)g(x)? don’t be surprised if your answer has a binomial coefficient in it.In fact, the binomial coefficient should help you finish the problem.

399. Since the sets of a k-set structure are nonempty and disjoint, the k-elementset of sets can be arranged as a k-tuple in k! ways.

403. The alternate definition of a funciton in Section 3.1.2 can be restated to saythat a function from a k-element set K to an n-element set N can be thought ofas an n-tuple of sets, perhaps with some empty, whose union is K. In order tothink of the function as an n-tuple, we number the elements of N as number1 through number n. Then the ith set in the n-tuple is the set of elementsmapped to the ith element of N in our numbering?

404. Don’t be surprised if you see a hyperbolic sine or hyperbolic cosine in youranswer. If you aren’t familiar with these functions, look them up in a calculusbook.

407. The EGF for∑n

i=1 (nk )k is

∑∞n=1

∑ni=1

n!k!(n−k)! k

xn

n! . You can cancel out the n!terms and the k terms. Now try to see if what is left can be regarded as theproduct of two EGFs.

421.a. To apply the exponential formula, we must take the exponential functionof an EGF whose constant term is zero, or in other words, for a species ofstructures that has no structures that use the empty set.

421.b. Once you know the vertex set of a graph, all you have to do to specify thegraph is to specify its set of edges.

421.d. What is the calculus definition of (1 + y)?

421.f. Look for a formula that involves summing over all partitions of the integer n.

Appendix E

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org/>Everyone is permitted to copy and distribute verbatim copies of this license document, but changing

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Index

S(k , n), 58Π notation, 8n!, 8

Stirling’s formula for, 19nk , 55nk , 8q-ary factorial, 84q-binomial coefficient, 84s(k , n), 62

action of a group on a set, 115arithmetic progression, 39arithmetic series, 40associative law, 105asymptotic combinatorics, 36automorphism (of a graph), 124, 131

basis (for polynomials), 61Bell Number, 59bĳection, 11, 137bĳection principle, 11binomial coefficient, 11

q-binomial, 84Binomial Theorem, 24binomial theorem

extended, 80block of a partition, 6, 141broken permutation, 57Burnside’s Lemma, 123

Cartesian product, 5Catalan Number, 22, 124

recurrence for, 89, 90Catalan number

generating function for, 89Catalan Path, 22

Cauchy-Frobenius-BurnsideTheorem, 123

characteristic function, 14chromatic polynomial of a graph, 99Chung-Feller Theorem, 23closure property, 105coefficient

multinomial, 60coloring

standard notation, 117standard ordering, 117

coloring of a graph, 98proper, 98

combinations, 11commutative law, 111complement, 96complement of a partition, 66composition, 28, 137

k parts, 28number of, 28

composition of functions, 104compositions

number of, 28congruence modulo n, 140conjugate of an integer partition, 65connected component graph, 165connected component of a graph, 98,

165connected structures and EGFs, 163constant coefficient linear

recurrence, 40contraction, 47cost of a spanning tree, 46cycle (in a graph), 43cycle (of a permutation), 112

element of, 112

199

200 Index

equivalent, 112cycle index, 129cycle monomial, 129cyclic group, 113

definitioninductive, 33recursive, 33

degree of a vertex, 42degree sequence, 50, 72

ordered, 50deletion, 47deletion-contraction recurrence, 48,

99derangement, 95derangement problem, 95diagram

of a partitionFerrers, 64Young, 64

digraph, 9, 135dihedral group, 107Dĳkstra’s algorithm, 48directed graph, 9, 135disjoint, 4

multisets, 121distance in a graph, 48distance in a weighted graph, 48domain (of a function), 133double induction, 35

strong, 35driving function, 40Dyck path, 22

edge, 26, 41, 135in a digraph, 135of a complete graph, 26

EGF, 152enumerator

fixed point, 127orbit, 126

equivalence class, 141equivalence relation, 140, 142equivalent cycle, 112exponential formula, 163

connected structures for, 165exponential generating function, 152exponential generating functions

product principle for, 160exponential generating functions for

connected structures, 163extended binomial theorem, 80

F-structures, 158factorial, 8, 33, 55

q-ary, 84falling, 55

factorial powerfalling, 8rising, 55

falling factorial power, 8, 55Ferrers diagram, 64Fibonacci numbers, 86, 88fix, 121fixed point enumerator, 127function, 3, 133

alternate definition, 55bĳection, 11characteristic, 14composition, 137digraph of, 9driving, 40identity, 104injection, 4inverse, 105one-to-one, 4, 134onto, 10, 134

and Stirling Numbers, 60ordered, 55

onto, 55relation of, 133surjection, 10, 134

functionscomposition of, 104number of, 53one-to-one

number of, 53onto

number of, 97

general product principle, 6, 34generating function, 76

exponential, 152product principle for, 160

ordinary, 152product principle for, 79

Index 201

geometric progression, 40geometric series, 40, 79graph, 41

chromatic polynomial of, 99coloring of, 98

proper, 98complete, 26connected component of, 98, 165coordinate, 135directed, 9, 135distance in, 48simple, 165

graphsisomorphic, 131

Gray Code, 28greedy method, 46group acting on a set, 115group action on colorings, 118group of permutations, 105

hatcheck problem, 95homogeneous linear recurrence, 40

identity function, 104, 137identity property, 105identity property (for permutations),

105inclusion and exclusion principle, 93

for unions of sets, 95indicator polynomials, 151induction

double, 35mathematical, the principle of,

31, 147mathematical, the strong

principle of, 32strong double, 35

inductiveconclusion, 32hypothesis, 32step, 32

inductive definition, 33injection, 4, 134inverse function, 105inverse property, 105involution, 114isometry, 108isomorphic

graphs, 131

k-set structures, 160

Lah number, 57lattice path, 20

diagonal, 20length (of a path), 48linear recurrence, 40, 87

constant coefficient, 40homogeneous, 40second order, 86

mathematical inductiondouble, 35principle of, 31, 147strong double, 35

ménage problem, 97method

probabilistic, 36minimum cost spanning tree, 46monochromatic subgraph, 37multinomial coefficient, 60multiorbit, 120multiorbits, 127multiplicity in a multiset, 56multiset, 56multisets

product principle, 122quotient principle, 122sum principle, 121union, 121

one-to-one, 4one-to-one function, 134onto function, 10, 134

counting, 60ordered, 55

onto functionsnumber of, 97

orbit, 119orbit enumerator, 126Orbit-Fixed Point Theorem, 127ordered degree sequence, 72ordered function, 55ordered onto function, 55ordered pair, 3ordinary generating function, 152

202 Index

pair structure, 159pair,ordered, 3partial fractions

method of, 88partition

blocks of, 6of a set, 5, 58

Stirling Numbers, 58partition (of a set), 141partition of a set

type vector, 59partition of an integer, 63

conjugate of, 65decreasing list, 64Ferrers diagram, 64into n parts, 63self conjugate, 65type vector, 64Young diagram, 64

partitions of a setnumber of, 59

Pascal’s Triangle, 12path

in graph, 43lattice, 20

diagonal, 20length of, 48

permutationk-element, 8as a bĳection, 11broken, 57cycle of, 112two row notation, 107

permutation group, 105picture enumerator, 74picture enumerators

product principle for, 75pigeonhole principle, 25

generalized, 26Pólya-Redfield Theorem, 129principle

bĳection, 11product, 5, 6

general, 6quotient, 143sum, 5, 6

principle of inclusion and exclusion,93

for unions of sets, 95principle of mathematical induction,

31, 147probabilistic method, 36product

Cartesian, 5product notation, 8product principle, 5, 6

for multisets, 122general, 6, 34picture enumerators, 75

product principle for exponentialgenerating functions, 160

product principle for generatingfunctions, 79

progressionarithmetic, 39geometric, 40

proper coloring of a graph, 98

quotient principle, 18, 143for multisets, 122

range (of a function), 133recurrence, 38

constant coefficient, 86, 87deletion-contraction, 48linear, 40, 86, 87linear homogeneous, 40second order, 86, 87solution to, 38two variable, 58

recurrence relation, 38recursive definition, 33reflexive, 139relation, 133

equivalence, 140, 142of a function, 133recurrence, 38reflexive, 139, 140transitive, 140

rising factorial power, 55rotation group, 105

second order recurrence, 86self-conjugate partition, 65

Index 203

sequencedegree, 50

seriesarithmetic, 40geometric, 40, 79

setcolorings of action of a group

on, 116sets

disjoint, 4mutually disjoint, 4

simple graph, 165space of polynomials, 61spanning tree, 45

cost of, 46minimum cost, 46

species, 158exponential generating function

for, 159standard notation for a coloring, 117Stirling Number

first kind, 62second kind, 58, 97

Stirling’s formula for n!, 19Stirling’s triangle

first kind, 62second kind, 58

strong double induction, 35strong principle of mathematical

induction, 32structure, 158

pair, 159

using a set, 158subgroup, 111sum principle, 5, 6, 93

for multisets, 121surjection, 10, 134surjections

number of, 97symmetric, 140symmetric group, 106

transitive, 140tree, 43

spanning, 45cost of, 46minimum cost, 46

Twentyfold Way, 52two row notation, 107type vector for a partition of an

integer, 64type vector of a partition of a set, 59

union of multisets, 121uses

a structure using a set, 158

vertex, 26, 41, 135degree of, 42of a complete graph, 26, 135

walk, 43

Young diagram, 64

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